Chapter 4 – Cubes and Cube Roots of RD Sharma Solutions for Class 8 Maths are provided here. RD Sharma Solutions help students secure good score in their examination, as they provide extensive knowledge about the subject, Class 8 being a critical stage in their academic career. BYJUâ€™S expert team has solved the questions from the RD Sharma Class 8 textbook in a step by step format in detail, which helps students strengthen their concepts.

Chapter 4- Cubes and Cube Roots contain five exercises and the RD Sharma Class 8 Solutions present in this page provides solutions for the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

- Cube of a number – A natural number is said to be a perfect cube if it is a cube of some natural number.
- Finding a cube of a two-digit number by column method.
- Cubes of Negative Integers.
- Cubes of Rational Numbers.
- Cube root of a Natural Number.
- Cube root of a negative integral perfect cube.
- Cube root of the product of integers.
- Finding cube roots using cube root tables.

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## EXERCISE 4.1 PAGE NO: 4.7

**1. Find the cubes of the following numbers:
(i) 7 (ii) 12 **

**(iii) 16 (iv) 21 **

**(v) 40 (vi) 55 **

**(vii) 100 (viii) 302 **

**(ix) 301**

**Solution:**

**(i)** 7

Cube of 7 is

7 = 7Ã— 7 Ã— 7 = 343

**(ii) **12

Cube of 12 is

12 = 12Ã— 12Ã— 12 = 1728

**(iii) **16

Cube of 16 is

16 = 16Ã— 16Ã— 16 = 4096

**(iv) **21

Cube of 21 is

21 = 21 Ã— 21 Ã— 21 = 9261

**(v) **40

Cube of 40 is

40 = 40Ã— 40Ã— 40 = 64000

**(vi)** 55

Cube of 55 is

55 = 55Ã— 55Ã— 55 = 166375

**(vii) **100

Cube of 100 is

100 = 100Ã— 100Ã— 100 = 1000000

**(viii) **302

Cube of 302 is

302 = 302Ã— 302Ã— 302 = 27543608

**(ix) **301

Cube of 301 is

301 = 301Ã— 301Ã— 301 = 27270901

**2.Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.**

**Solutions:**

Firstly let us find the Cube of natural numbers up to 10

1^{3}Â = 1 Ã— 1 Ã— 1 = 1

2^{3}Â = 2 Ã— 2 Ã— 2 = 8

3^{3}Â = 3 Ã— 3 Ã— 3 = 27

4^{3}Â = 4 Ã— 4 Ã— 4 = 64

5^{3}Â = 5 Ã— 5 Ã— 5 = 125

6^{3}Â = 6 Ã— 6 Ã— 6 = 216

7^{3}Â = 7 Ã— 7 Ã— 7 = 343

8^{3}Â = 8 Ã— 8 Ã— 8 = 512

9^{3}Â = 9 Ã— 9 Ã— 9 = 729

10^{3} = 10 Ã— 10 Ã— 10 = 1000

âˆ´ From the above results we can say that

(i) Cubes of all odd natural numbers are odd.

(ii) Cubes of all even natural numbers are even.

**3. Observe the following pattern:
1 ^{3} = 1**

**1 ^{3} + 2^{3} = (1+2)^{2}**

**1 ^{3} + 2^{3} + 3^{3} = (1+2+3)^{2}**

Write the next three rows and calculate the value of 1^{3} + 2^{3} + 3^{3} +â€¦+ 9^{3}Â by the above pattern.

**Solution:**

According to given pattern,

1^{3} + 2^{3} + 3^{3} +â€¦+ 9^{3}

1^{3} + 2^{3} + 3^{3} +â€¦+ n^{3}Â = (1+2+3+â€¦+n)^{ 2}

So when n = 10

1^{3} + 2^{3} + 3^{3} +â€¦+ 9^{3 }+ 10^{3}Â = (1+2+3+â€¦+10)^{ 2}

= (55)^{2} = 55Ã—55 = 3025

**4. Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
â€œThe cube of a natural number which is a multiple of 3 is a multiple of 27â€™**

**Solution:**

We know that the first 5 natural numbers which are multiple of 3 are 3, 6, 9, 12 and 15

So now, let us find the cube of 3, 6, 9, 12 and 15

3^{3}Â = 3 Ã— 3 Ã— 3 = 27

6^{3}Â = 6 Ã— 6 Ã— 6 = 216

9^{3}Â = 9 Ã— 9 Ã— 9 = 729

12^{3}Â = 12 Ã— 12 Ã— 12 = 1728

15^{3}Â = 15 Ã— 15 Ã— 15 = 3375

We found that all the cubes are divisible by 27

âˆ´ â€œThe cube of a natural number which is a multiple of 3 is a multiple of 27â€™

**5.Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, â€¦)Â and verify the following:
â€œThe cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1â€™**

**Solution:**

We know that the first 5 natural numbers in the form of (3n + 1) are 4, 7, 10, 13 and 16

So now, let us find the cube of 4, 7, 10, 13 and 16

4^{3}Â = 4 Ã— 4 Ã— 4 = 64

7^{3}Â = 7 Ã— 7 Ã— 7 = 343

10^{3}Â = 10 Ã— 10 Ã— 10 = 1000

13^{3}Â = 13 Ã— 13 Ã— 13 = 2197

16^{3}Â = 16 Ã— 16 Ã— 16 = 4096

We found that all these cubeswhen divided by â€˜3â€™ leaves remainder 1.

âˆ´ the statement â€œThe cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1â€™ is true.

**6. Write the cubes 5 natural numbers of the from 3n+2(i.e.5,8,11â€¦.) and verify the following:
â€œThe cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2â€™**

**Solution**:

We know that the first 5 natural numbers in the form (3n + 2) are 5, 8, 11, 14 and 17

So now, let us find the cubes of 5, 8, 11, 14 and 17

5^{3}Â = 5 Ã— 5 Ã— 5 = 125

8^{3}Â = 8 Ã— 8 Ã— 8 = 512

11^{3}Â = 11 Ã— 11 Ã— 11 = 1331

14^{3}Â = 14 Ã— 14 Ã— 14 = 2744

17^{3}Â = 17 Ã— 17 Ã— 17 = 4313

We found that all these cubes when divided by â€˜3â€™ leaves remainder 2.

âˆ´ thestatementâ€œThe cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2â€™ is true.

**7.Write the cubes of 5 natural numbers of which are multiples of 7 and verity the following:
â€œThe cube of a multiple of 7 is a multiple of 7 ^{3}.**

**Solution:**

The first 5 natural numbers which are multiple of 7 are 7, 14, 21, 28 and 35

So, the Cube of 7, 14, 21, 28 and 35

7^{3}Â = 7 Ã— 7 Ã— 7 = 343

14^{3}Â = 14 Ã— 14 Ã— 14 = 2744

21^{3}Â = 21Ã— 21Ã— 21 = 9261

28^{3}Â = 28 Ã— 28 Ã— 28 = 21952

35^{3}Â = 35 Ã— 35 Ã— 35 = 42875

We found that all these cubes are multiples of 7^{3}(343) as well.

âˆ´The statementâ€œThe cube of a multiple of 7 is a multiple of 7^{3} is true.

**8. Which of the following are perfect cubes?
(i) 64 (ii) 216
(iii) 243 (iv) 1000
(v) 1728 (vi) 3087
(vii) 4608 (viii) 106480
(ix) 166375 (x) 456533**

**Solution:**

**(i)** 64

First find the factors of 64

64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 2^{6}Â = (2^{2})^{3}Â = 4^{3}

Hence, itâ€™s a perfect cube.

**(ii)** 216

First find thefactors of 216

216 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 = 2^{3}Â Ã— 3^{3}Â = 6^{3}

Hence, itâ€™s a perfect cube.

**(iii)** 243

First find thefactors of 243

243 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 = 3^{5}Â = 3^{3}Â Ã— 3^{2}

Hence, itâ€™s not a perfect cube.

**(iv)** 1000

First find thefactors of 1000

1000 = 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5 = 2^{3}Â Ã— 5^{3}Â = 10^{3}

Hence, itâ€™s a perfect cube.

**(v)** 1728

First find thefactors of 1728

1728 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 = 2^{6}Â Ã— 3^{3}Â = (4 Ã— 3 )^{3}Â = 12^{3}

Hence, itâ€™s a perfect cube.

**(vi)** 3087

First find thefactors of 3087

3087 = 3 Ã— 3 Ã— 7 Ã— 7 Ã— 7 = 3^{2}Â Ã— 7^{3}

Hence, itâ€™s not a perfect cube.

**(vii)** 4608

First find thefactors of 4608

4608 = 2 Ã— 2 Ã— 3 Ã— 113

Hence, itâ€™s not a perfect cube.

**(viii)** 106480

First find thefactors of 106480

106480 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5 Ã— 11 Ã— 11 Ã— 11

Hence, itâ€™s not a perfect cube.

**(ix)** 166375

First find thefactors of 166375

166375= 5 Ã— 5 Ã— 5 Ã— 11 Ã— 11 Ã— 11 = 5^{3}Â Ã— 11^{3}Â = 55^{3}

Hence, itâ€™s a perfect cube.

**(x)** 456533

First find thefactors of 456533

456533= 11 Ã— 11 Ã— 11 Ã— 7 Ã— 7 Ã— 7 = 11^{3}Â Ã— 7^{3}Â = 77^{3}

Hence, itâ€™s a perfect cube.

**9. Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824**

**Solution:**

**(i)** 216 = 2^{3}Â Ã— 3^{3}Â = 6^{3}

Itâ€™s a cube of even natural number.

**(ii)** 512 = 2^{9}Â = (2^{3})^{3}Â = 8^{3}

Itâ€™s a cube of even natural number.

**(iii)** 729 = 3^{3}Â Ã— 3^{3}Â = 9^{3}

Itâ€™s not a cube of even natural number.

**(iv)** 1000 = 10^{3}

Itâ€™s a cube of even natural number.

**(v)** 3375 = 3^{3}Â Ã— 5^{3}Â = 15^{3}

Itâ€™s not a cube of even natural number.

**(vi)** 13824 = 2^{9 }Ã— 3^{3} = (2^{3})^{3}Â Ã— 3^{3} = 8^{3}Ã—3^{3 }= 24^{3}

Itâ€™s a cube of even natural number.

**10. Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859**

**Solution:**

**(i)** 125 = 5 Ã— 5 Ã— 5 Ã— 5 = 5^{3}

Itâ€™s a cube of odd natural number.

**(ii)** 343 = 7 Ã— 7 Ã— 7 = 7^{3}

Itâ€™s a cube of odd natural number.

**(iii)** 1728 = 2^{6}Â Ã— 3^{3}Â = 4^{3}Â Ã— 3^{3}Â = 12^{3}

Itâ€™s not a cube of odd natural number. As 12 is even number.

**(iv)** 4096 = 2^{12}Â = (2^{6})^{2}Â = 64^{2}

Its not a cube of odd natural number. As 64 is an even number.

**(v)** 32768 = 2^{15}Â = (2^{5})^{3}Â = 32^{3}

Itâ€™s not a cube of odd natural number. As 32 is an even number.

**(vi)** 6859 = 19 Ã— 19 Ã— 19 = 19^{3}

Itâ€™s a cube of odd natural number.

**11. What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?
(i) 675 (ii) 1323
(iii) 2560 (iv) 7803
(v) 107811 (vi) 35721**

**Solution:**

**(i)** 675

First find the factors of 675

675 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5

= 3^{3}Â Ã— 5^{2}

âˆ´To make a perfect cube we need to multiply the product by 5.

**(ii)** 1323

First find the factors of 1323

1323 = 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7

= 3^{3}Â Ã— 7^{2}

âˆ´To make a perfect cube we need to multiply the product by 7.

**(iii)** 2560

First find the factors of 2560

2560 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5

= 2^{3}Â Ã— 2^{3}Â Ã— 2^{3}Â Ã— 5

âˆ´To make a perfect cube we need to multiply the product by 5 Ã— 5 = 25.

**(iv)** 7803

First find the factors of 7803

7803 = 3 Ã— 3 Ã— 3 Ã— 17 Ã— 17

= 3^{3}Â Ã— 17^{2}

âˆ´To make a perfect cube we need to multiply the product by 17.

**(v)** 107811

First find the factors of 107811

107811 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 11 Ã— 11 Ã— 11

= 3^{3}Â Ã— 3 Ã— 11^{3}

âˆ´To make a perfect cube we need to multiply the product by 3 Ã— 3 = 9.

**(vi)** 35721

First find the factors of 35721

35721 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7

= 3^{3}Â Ã— 3^{3}Â Ã— 7^{2}

âˆ´To make a perfect cube we need to multiply the product by 7.

**12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
(i) 675 (ii) 8640
(iii) 1600 (iv) 8788
(v) 7803 (vi) 107811
(vii) 35721 (viii) 243000**

**Solution:**

**(i)** 675

First find the prime factors of 675

675 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5

= 3^{3}Â Ã— 5^{2}

Since 675 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5^{2}Â = 25, which gives 27 as quotient where, 27 is a perfect cube.

âˆ´ 25 is the required smallest number.

**(ii)** 8640

First find the prime factors of 8640

8640 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 5

= 2^{3}Â Ã— 2^{3}Â Ã— 3^{3}Â Ã— 5

Since 8640 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5, which gives 1728 as quotient and we know that 1728 is a perfect cube.

âˆ´5 is the required smallest number.

**(iii)** 1600

First find the prime factors of 1600

1600 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5

= 2^{3}Â Ã— 2^{3}Â Ã— 5^{2}

Since 1600 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5^{2}Â = 25, which gives 64 as quotient and we know that 64 is a perfect cube

âˆ´ 25 is the required smallest number.

**(iv)** 8788

First find the prime factors of 8788

8788 = 2 Ã— 2 Ã— 13 Ã— 13 Ã— 13

= 2^{2}Â Ã— 13^{3}

Since 8788 is not a perfect cube.

To make the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and we know that 2197 is a perfect cube

âˆ´ 4 is the required smallest number.

**(v)** 7803

First find the prime factors of 7803

7803 = 3 Ã— 3 Ã— 3 Ã— 17 Ã— 17

= 3^{3}Â Ã— 17^{2}

Since 7803 is not a perfect cube.

To make the quotient a perfect cube we divide it by 17^{2}Â = 289, which gives 27 as quotient and we know that 27 is a perfect cube

âˆ´ 289 is the required smallest number.

**(vi)** 107811

First find the prime factors of 107811

107811 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 11 Ã— 11 Ã— 11

= 3^{3}Â Ã— 11^{3}Â Ã— 3

Since 107811 is not a perfect cube.

To make the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube.

âˆ´ 3 is the required smallest number.

**(vii)** 35721

First find the prime factors of 35721

35721 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7

= 3^{3}Â Ã— 3^{3}Â Ã— 7^{2}

Since 35721 is not a perfect cube.

To make the quotient a perfect cube we divide it by 7^{2}Â = 49, which gives 729 as quotient and we know that 729 is a perfect cube

âˆ´ 49 is the required smallest number.

**(viii)** 243000

First find the prime factors of 243000

243000 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 5

= 2^{3}Â Ã— 3^{3}Â Ã— 5^{3}Â Ã— 3^{2}

Since 243000 is not a perfect cube.

To make the quotient a perfect cube we divide it by 3^{2}Â = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube

âˆ´ 9 is the required smallest number.

**13. Prove that if a number is trebled then its cube is 27 time the cube of the given number.**

**Solution:**

Let us consider a number as a

So the cube of the assumed number is = a^{3}

Now, the number is trebled = 3 Ã— a = 3a

So the cube of new number = (3a)^{ 3}Â = 27a^{3}

âˆ´New cube is 27 times of the original cube.

Hence, proved.

**14. What happens to the cube of a number if the number is multiplied by
(i) 3?
(ii) 4?
(iii) 5?**

**Solution:**

**(i)** 3?

Let us consider the number as a

So its cube will be = a^{3}

According to the question, the number is multiplied by 3

New number becomes = 3a

So the cube of new number will be = (3a)^{ 3}Â = 27a^{3}

Hence, number will becomeÂ 27 times the cube of the number.

**(ii)** 4?

Let us consider the number as a

So its cube will be = a^{3}

According to the question, the number is multiplied by 4

New number becomes = 4a

So the cube of new number will be = (4a)^{ 3}Â = 64a^{3}

Hence, number will becomeÂ 64 times the cube of the number.

**(iii)** 5?

Let us consider the number as a

So its cube will be = a^{3}

According to the question, the number is multiplied by 5

New number becomes = 5a

So the cube of new number will be = (5a)^{ 3}Â = 125a^{3}

Hence, number will becomeÂ 125 times the cube of the number.

**15. Find the volume of a cube, one face of which has an area of 64m ^{2}.**

**Solution:**

We know that the given area of one face of cube = 64 m^{2}

Let the length of edge of cube be â€˜aâ€™ metre

a^{2} = 64

a = âˆšÂ 64

= 8m

Now, volume of cube = a^{3}

a^{3} = 8^{3} = 8 Ã— 8 Ã— 8

= 512m^{3}

âˆ´Volume of a cube is 512m^{3}

**16. Find the volume of a cube whose surface area is 384m ^{2}.**

**Solution:**

We know that the surface area of cube = 384 m^{2}

Let us consider the length of each edge of cube be â€˜aâ€™ meter

6a^{2} = 384

a^{2} = 384/6

= 64

a = âˆš64

= 8m

Now, volume of cube = a^{3}

a^{3} = 8^{3} = 8 Ã— 8 Ã— 8

= 512m^{3}

âˆ´ Volume of a cube is 512m^{3}

**17. Evaluate the following:
(i)Â {(5 ^{2} + 12^{2})^{1/2}}^{3}
(ii)Â {(6^{2} + 8^{2})^{1/2}}^{3}**

**Solution:**

(i)Â {(5^{2} + 12^{2})^{1/2}}^{3}

When simplified above equation we get,

{(25 + 144)^{1/2}}^{3}

{(169)^{1/2}}^{3}

{(13^{2})^{1/2}}^{3}

(13)^{3}

2197

(ii)Â {(6^{2} + 8^{2})^{1/2}}^{3}

When simplified above equation we get,

{(36 + 64)^{1/2}}^{3}

{(100)^{1/2}}^{3}

{(10^{2})^{1/2}}^{3}

(10)^{3}

1000

**18. Write the units digit of the cube of each of the following numbers:
31, 109, 388, 4276, 5922, 77774, 44447, 125125125**

**Solution:**

**31 **

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 31 is 1

Cube of 1 = 1^{3}Â = 1

âˆ´ Unit digit of cube of 31 is always 1

**109**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 109 is = 9

Cube of 9 = 9^{3}Â = 729

âˆ´ Unit digit of cube of 109 is always 9

**388**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 388 is = 8

Cube of 8 = 8^{3}Â = 512

âˆ´ Unit digit of cube of 388 is always 2

**4276**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 4276 is = 6

Cube of 6 = 6^{3}Â = 216

âˆ´ Unit digit of cube of 4276 is always 6

**5922**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 5922 is = 2

Cube of 2 = 2^{3}Â = 8

âˆ´ Unit digit of cube of 5922 is always 8

**77774**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 77774 is = 4

Cube of 4 = 4^{3}Â = 64

âˆ´ Unit digit of cube of 77774 is always 4

**44447**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 44447 is = 7

Cube of 7 = 7^{3}Â = 343

âˆ´ Unit digit of cube of 44447 is always 3

**125125125**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 125125125 is = 5

Cube of 5 = 5^{3}Â = 125

âˆ´ Unit digit of cube of 125125125 is always 5

**19. Find the cubes of the following numbers by column method:
(i) 35
(ii) 56
(iii) 72**

**Solution:**

**(i)** 35

We have, a = 3 and b = 5

Column I
a |
Column II
3Ã—a |
Column III
3Ã—aÃ—b |
Column IV
b |

3^{3} = 27 |
3Ã—9Ã—5 = 135 | 3Ã—3Ã—25 = 225 | 5^{3} = 125 |

+15 | +23 | +12 | 125 |

42 | 158 | 237 | |

42 | 8 | 7 | 5 |

âˆ´ The cube of 35 is 42875

**(ii) **56

We have, a = 5 and b = 6

Column I
a |
Column II
3Ã—a |
Column III
3Ã—aÃ—b |
Column IV
b |

5^{3} = 125 |
3Ã—25Ã—6 = 450 | 3Ã—5Ã—36 = 540 | 6^{3} = 216 |

+50 | +56 | +21 | 126 |

175 | 506 | 561 | |

175 | 6 | 1 | 6 |

âˆ´ The cube of 56 is 175616

**(iii) 72**

We have, a = 7 and b = 2

Column I
a |
Column II
3Ã—a |
Column III
3Ã—aÃ—b |
Column IV
b |

7^{3} = 343 |
3Ã—49Ã—2 = 294 | 3Ã—7Ã—4 = 84 | 2^{3} = 8 |

+30 | +8 | +0 | 8 |

373 | 302 | 84 | |

373 | 2 | 4 | 8 |

âˆ´ The cube of 72 is 373248

**20. Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728**

**Solution:**

(i) 64

Firstly let us find the prime factors of 64

64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 2^{3}Â Ã— 2^{3}

= 4^{3}

Hence, itâ€™s a perfect cube.

(ii) 216

Firstly let us find the prime factors of 216

216 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 2^{3}Â Ã— 3^{3}

= 6^{3}

Hence, itâ€™s a perfect cube.

(iii) 243

Firstly let us find the prime factors of 243

243 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 3^{3}Â Ã— 3^{2}

Hence, itâ€™s not a perfect cube.

(iv) 1728

Firstly let us find the prime factors of 1728

1728 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 2^{3}Â Ã— 2^{3}Â Ã— 3^{3}

= 12^{3}

Hence, itâ€™s a perfect cube.

**21. For each of the non-perfect cubes in Q. No 20 find the smallest number by which it must be
(a) Multiplied so that the product is a perfect cube.
(b) Divided so that the quotient is a perfect cube.**

**Solution:**

Only non-perfect cube in previous question was = 243

**(a)** Multiplied so that the product is a perfect cube.

Firstly let us find the prime factors of 243

243 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 = 3^{3}Â Ã— 3^{2}

Hence, to make it a perfect cube we should multiply it by 3.

**(b)** Divided so that the quotient is a perfect cube.

Firstly let us find the prime factors of 243

243 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 = 3^{3}Â Ã— 3^{2}

Hence, to make it a perfect cube we have to divide it by 9.

**22. By taking three different, values of n verify the truth of the following statements:
(i) If n is even, then n ^{3}Â is also even.
(ii) If n is odd, then n^{3}Â is also odd.
(ii) If n leaves remainder 1 when divided by 3, then n^{3}Â also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p+2 then n^{3}Â also a number of the same type.**

**Solution:**

**(i)** If n is even, then n^{3}Â is also even.

Let us consider three even natural numbers 2, 4, 6

So now, Cubes of 2, 4 and 6 are

2^{3}Â = 8

4^{3}Â = 64

6^{3}Â = 216

Hence, we can see that all cubes are even in nature.

Statement is verified.

**(ii)** If n is odd, then n^{3}Â is also odd.

Let us consider three odd natural numbers 3, 5, 7

So now, cubes of 3, 5 and 7 are

3^{3}Â = 27

5^{3}Â = 125

7^{3}Â = 343

Hence, we can see that all cubes are odd in nature.

Statement is verified.

**(iii)** If n leaves remainder 1 when divided by 3, then n^{3}Â also leaves 1 as remainder when divided by 3.

Let us consider three natural numbers of the form (3n+1) are 4, 7 and 10

So now, cube of 4, 7, 10 are

4^{3}Â = 64

7^{3}Â = 343

10^{3}Â = 1000

We can see that if we divide these numbers by 3, we get 1 as remainder in each case.

Hence, statement is verified.

**(iv)** If a natural number n is of the form 3p+2 then n^{3}Â also a number of the same type.

Let us consider three natural numbers of the form (3p+2) are 5, 8 and 11

So now, cube of 5, 8 and 10 are

5^{3}Â = 125

8^{3}Â = 512

11^{3}Â = 1331

Now, we try to write these cubes in form of (3p + 2)

125 = 3 Ã— 41 + 2

512 = 3 Ã— 170 + 2

1331 = 3 Ã— 443 + 2

Hence, statement is verified.

**23. Write true (T) or false (F) for the following statements:
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a**

^{3}Â is always greater than a

^{2}. (vi) If a and b are integers such that a

^{2}>b

^{2}, then a

^{3}>b

^{3}. (vii) If a divides b, then a

^{3}Â divides b

^{3}. (viii) If a

^{2}Â ends in 9, then a

^{3}Â ends in 7. (ix) If a

^{2}Â ends in an even number of zeros, then a

^{3}Â ends in 25. (x) If a

^{2}Â ends in an even number of zeros, then a

^{3}Â ends in an odd number of zeros.

**Solution:**

(i) 392 is a perfect cube.

Firstly letâ€™s find the prime factors of 392 = 2 Ã— 2 Ã— 2 Ã— 7 Ã— 7 = 2^{3}Â Ã— 7^{2}

Hence the statement is False.

(ii) 8640 is not a perfect cube.

Prime factors of 8640 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 5 = 2^{3}Â Ã— 2^{3}Â Ã— 3^{3}Â Ã— 5

Hence the statement is True

(iii) No cube can end with exactly two zeros.

Statement is True.

Because a perfect cube always have zeros in multiple of 3.

(iv) There is no perfect cube which ends in 4.

We know 64 is a perfect cube = 4 Ã— 4 Ã— 4 and it ends with 4.

Hence the statement is False.

(v) For an integer a, a^{3}Â is always greater than a^{2}.

Statement is False.

Because in case of negative integers ,

(-2)^{2} = 4 and (-2)^{3} = -8

(vi) If a and b are integers such that a^{2}>b^{2}, then a^{3}>b^{3}.

Statement is False.

In case of negative integers,

(-5)^{2} > (-4)^{2} = 25 > 16

But, (-5)^{3} > (-4)^{3} = -125 > -64 is not true.

(vii) If a divides b, then a^{3}Â divides b^{3}.

Statement is True.

If a divides b

b/a = k, so b=ak

b^{3}/a^{3} = (ak)^{3}/a^{3} = a^{3}k^{3}/a^{3} = k^{3},

For each value of b and a its true.

(viii) If a^{2}Â ends in 9, then a^{3}Â ends in 7.

Statement is False.

Let a = 7

7^{2}Â = 49 and 7^{3}Â = 343

(ix) If a^{2}Â ends in an even number of zeros, then a^{3}Â ends in 25.

Statement is False.

Since, when a = 20

a^{2}Â = 20^{2}Â = 400 and a^{3}Â = 8000 (a^{3} doesnâ€™t end with 25)

(x) If a^{2}Â ends in an even number of zeros, then a^{3}Â ends in an odd number of zeros.

Statement is False.

Since, when a = 100

a^{2}Â = 100^{2}Â = 10000 and a^{3}Â = 100^{3}Â = 1000000 (a^{3} doesnâ€™t end with odd number of zeros)

## EXERCISE 4.2 PAGE NO: 4.13

**1. Find the cubes of:
(i) -11
(ii) -12
(iii) -21**

**Solution:**

(i) -11

The cube of 11 is

(-11)^{3} = -11Ã— -11Ã— -11 = -1331

(ii) -12

The cube of 12 is

(-12)^{3} = -12Ã— -12Ã— -12 = -1728

(iii) -21

The cube of 21 is

(-21)^{3} = -21Ã— -21Ã— -21 = -9261

**2. Which of the following integers are cubes of negative integers
(i) -64
(ii) -1056
(iii) -2197
(iv) -2744
(v) -42875**

**Solution:**

**(i)** -64

The prime factors of 64 are

64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 2^{3}Â Ã— 2^{3}

= 4^{3}

âˆ´ 64 is a perfect cube of negative integer â€“ 4.

**(ii)** -1056

The prime factors of 1056 are

1056 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 11

1056 is not a perfect cube.

âˆ´ -1056 is not a cube of negative integer.

**(iii)** -2197

The prime factors of 2197 are

2197 = 13 Ã— 13 Ã— 13

= 13^{3}

âˆ´ 2197 is a perfect cube of negative integer â€“ 13.

**(iv)** -2744

The prime factors of 2744 are

2744 = 2 Ã— 2 Ã— 2 Ã— 7 Ã— 7 Ã— 7

= 2^{3}Â Ã— 7^{3}

= 14^{3}

2744 is a perfect cube.

âˆ´ -2744 is a cube of negative integer â€“ 14.

**(v)** -42875

The prime factors of 42875 are

42875 = 5 Ã— 5 Ã— 5 Ã— 7 Ã— 7 Ã— 7

= 5^{3}Â Ã— 7^{3}

= 35^{3}

42875 is a perfect cube.

âˆ´ -42875 is a cube of negative integer â€“ 35.

**3. Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer.
(i) -5832
(ii) -2744000**

**Solution:**

**(i)** -5832

The prime factors of 5832 are

5823 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 2^{3}Â Ã— 3^{3}Â Ã— 3^{3}

= 18^{3}

5832 is a perfect cube.

âˆ´ -5832 is a cube of negative integer â€“ 18.

**(ii)** -2744000

The prime factors of 2744000 are

2744000 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5 Ã— 7 Ã— 7 Ã— 7

= 2^{3}Â Ã— 2^{3}Ã— 5^{3}Â Ã— 7^{3}

2744000 is a perfect cube.

âˆ´ -2744000 is a cube of negative integer â€“ 140.

**4. Find the cube of:
(i)Â 7/9Â (ii)Â -8/11
(iii)Â 12/7Â (iv)Â -13/8
(v)Â 2 2/5Â (vi)Â 3 1/4
(vii) 0.3 (viii) 1.5
(ix) 0.08 (x) 2.1**

**Solution:**

**(i)**Â 7/9

The cube of 7/9 is

(7/9)^{3} = 7^{3}/9^{3} = 343/729

**(ii)**Â -8/11

The cube of -8/11 is

(-8/11)^{3} = -8^{3}/11^{3} = -512/1331

**(iii)**Â 12/7

The cube of 12/7 is

(12/7)^{3} = 12^{3}/7^{3} = 1728/343

**(iv)**Â -13/8

The cube of -13/8 is

(-13/8)^{3} = -13^{3}/8^{3} = -2197/512

**(v)**Â 2 2/5

The cube of 12/5 is

(12/5)^{3} = 12^{3}/5^{3} = 1728/125

**(vi)**Â 3 Â¼

The cube of 13/4 is

(13/4)^{3} = 13^{3}/4^{3} = 2197/64

**(vii)** 0.3

The cube of 0.3 is

(0.3)^{3} = 0.3Ã—0.3Ã—0.3 = 0.027

**(viii)** 1.5

The cube of 1.5 is

(1.5)^{3} = 1.5Ã—1.5Ã—1.5 = 3.375

**(ix)** 0.08

The cube of 0.08 is

(0.08)^{3} = 0.08Ã—0.08Ã—0.08 = 0.000512

**(x)** 2.1

The cube of 2.1 is

(2.1)^{3} = 2.1Ã—2.1Ã—2.1 = 9.261

**5. Find which of the following numbers are cubes of rational numbers:
(i)Â 27/64
(ii)Â 125/128
(iii) 0.001331
(iv) 0.04**

**Solution:**

**(i)**Â 27/64

We have,

27/64 = (3Ã—3Ã—3)/ (4Ã—4Ã—4) = 3^{3}/4^{3} = (3/4)^{3}

âˆ´ 27/64 is a cube of 3/4.

**(ii)**Â 125/128

We have,

125/128 = (5Ã—5Ã—5)/ (2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2) = 5^{3}/ (2^{3}Ã—2^{3}Ã—2)

âˆ´ 125/128 is not a perfect cube.

**(iii)** 0.001331

We have,

1331/1000000 = (11Ã—11Ã—11)/ (100Ã—100Ã—100) = 11^{3}/100^{3} = (11/100)^{3}

âˆ´ 0.001331 is a perfect cube ofÂ 11/100

**(iv)** 0.04

We have,

4/10 = (2Ã—2)/(2Ã—5) = 2^{2}/(2Ã—5)

âˆ´ 0.04 is not a perfect cube.

## EXERCISE 4.3 PAGE NO: 4.21

**1. Find the cube roots of the following numbers by successive subtraction of numbers:
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, â€¦
(i) 64
(ii) 512
(iii) 1728**

**Solution:**

**(i)** 64

Letâ€™s perform subtraction

64 â€“ 1 = 63

63 â€“ 7 = 56

56 â€“ 19 =37

37 â€“ 37 = 0

Subtraction is performed 4 times.

âˆ´ Cube root of 64 is 4.

**(ii)** 512

Letâ€™s perform subtraction

512 â€“ 1 = 511

511 â€“ 7 = 504

504 â€“ 19 = 485

485 â€“ 37 = 448

448 â€“ 61 = 387

387 â€“ 91 = 296

296 â€“ 127 = 169

169 â€“ 169 = 0

Subtraction is performed 8 times.

âˆ´ Cube root of 512 is 8.

**(iii)** 1728

Letâ€™s perform subtraction

1728 â€“ 1 = 1727

1727 â€“ 7 = 1720

1720 â€“ 19 = 1701

1701 â€“ 37 = 1664

1664 â€“ 91 = 1512

1512 â€“ 127 = 1385

1385 â€“ 169 = 1216

1216 â€“ 217 = 999

999 â€“ 271 = 728

728 â€“ 331 = 397

397 â€“ 397 = 0

Subtraction is performed 12 times.

âˆ´ Cube root of 1728 is 12.

**2. Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:
(i) 130
(ii) 345
(iii) 792
(iv) 1331**

**Solution:**

**(i)** 130

Letâ€™s perform subtraction

130 â€“ 1 = 129

129 â€“ 7 = 122

122 â€“ 19 = 103

103 â€“ 37 = 66

66 â€“ 61 = 5

Next number to be subtracted is 91, which is greater than 5

âˆ´130 is not a perfect cube.

**(ii)** 345

Letâ€™s perform subtraction

345 â€“ 1 = 344

344 â€“ 7 = 337

337 â€“ 19 = 318

318 â€“ 37 = 281

281 â€“ 61 = 220

220 â€“ 91 = 129

129 â€“ 127 = 2

Next number to be subtracted is 169, which is greater than 2

âˆ´ 345 is not a perfect cube

**(iii)** 792

Letâ€™s perform subtraction

792 â€“ 1 = 791

791 â€“ 7 = 784

784 â€“ 19 = 765

765 â€“ 37 = 728

728 â€“ 61 = 667

667 â€“ 91 = 576

576 â€“ 127 = 449

449 â€“ 169 = 280

280 â€“ 217 = 63

Next number to be subtracted is 271, which is greater than 63

âˆ´ 792 is not a perfect cube

**(iv)** 1331

Letâ€™s perform subtraction

1331 â€“ 1 = 1330

1330 â€“ 7 = 1323

1323 â€“ 19 = 1304

1304 â€“ 37 = 1267

1267 â€“ 61 = 1206

1206 â€“ 91 = 1115

1115 â€“ 127 = 988

988 â€“ 169 = 819

819 â€“ 217 = 602

602 â€“ 271 = 331

331 â€“ 331 = 0

Subtraction is performed 11 times

Cube root of 1331 is 11

âˆ´ 1331 is a perfect cube.

**3. Find the smallest number that must be subtracted from those of the numbers in question 2 which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots?**

**Solution:**

In previous question there are three numbers which are not perfect cubes.

**(i)** 130

Letâ€™s perform subtraction

130 â€“ 1 = 129

129 â€“ 7 = 122

122 â€“ 19 = 103

103 â€“ 37 = 66

66 â€“ 61 = 5

Next number to be subtracted is 91, which is greater than 5

Since, 130 is not a perfect cube. So, to make it perfect cube we subtract 5 from the given number.

130 â€“ 5 = 125 (which is a perfect cube of 5)

**(ii)** 345

Letâ€™s perform subtraction

345 â€“ 1 = 344

344 â€“ 7 = 337

337 â€“ 19 = 318

318 â€“ 37 = 281

281 â€“ 61 = 220

220 â€“ 91 = 129

129 â€“ 127 = 2

Next number to be subtracted is 169, which is greater than 2

Since, 345 is not a perfect cube. So, to make it a perfect cube we subtract 2 from the given number.

345 â€“ 2 = 343 (which is a perfect cube of 7)

**(iii)** 792

Letâ€™s perform subtraction

792 â€“ 1 = 791

791 â€“ 7 = 784

784 â€“ 19 = 765

765 â€“ 37 = 728

728 â€“ 61 = 667

667 â€“ 91 = 576

576 â€“ 127 = 449

449 â€“ 169 = 280

280 â€“ 217 = 63

Next number to be subtracted is 271, which is greater than 63

Since, 792 is not a perfect cube. So, to make it a perfect cube we subtract 63 from the given number.

792 â€“ 63 = 729 (which is a perfect cube of 9)

**4. Find the cube root of each of the following natural numbers:
(i) 343 (ii) 2744**

**(iii) 4913 (iv) 1728
(v) 35937 (vi) 17576
(vii) 134217728 (viii) 48228544
(ix) 74088000 (x) 157464
(xi) 1157625 (xii) 33698267**

**Solution:**

**(i)** 343

By using prime factorization method

âˆ›343 = âˆ› (7Ã—7Ã—7) = 7

**(ii)** 2744

By using prime factorization method

âˆ›2744 = âˆ› (2Ã—2Ã—2Ã—7Ã—7Ã—7) = âˆ› (2^{3}Ã—7^{3}) = 2Ã—7 = 14

**(iii)** 4913

By using prime factorization method,

âˆ›4913 = âˆ› (17Ã—17Ã—17) = 17

**(iv)** 1728

By using prime factorization method,

âˆ›1728 = âˆ›(2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3) = âˆ› (2^{3}Ã—2^{3}Ã—3^{3}) = 2Ã—2Ã—3 = 12

**(v)** 35937

By using prime factorization method,

âˆ›35937 = âˆ› (3Ã—3Ã—3Ã—11Ã—11Ã—11) = âˆ› (3^{3}Ã—11^{3}) = 3Ã—11 = 33

**(vi)** 17576

By using prime factorization method,

âˆ›17576 = âˆ› (2Ã—2Ã—2Ã—13Ã—13Ã—13) = âˆ› (2^{3}Ã—13^{3}) = 2Ã—13 = 26

**(vii)** 134217728

By using prime factorization method

âˆ›134217728 = âˆ› (2^{27}) = 2^{9} = 512

**(viii)** 48228544

By using prime factorization method

âˆ›48228544 = âˆ› (2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—7Ã—7Ã—7Ã—13Ã—13Ã—13) = âˆ› (2^{3}Ã—2^{3}Ã—7^{3}Ã—13^{3}) = 2Ã—2Ã—7Ã—13 = 364

**(ix)** 74088000

By using prime factorization method

âˆ›74088000 = âˆ› (2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7) = âˆ› (2^{3}Ã—2^{3}Ã—3^{3}Ã—5^{3}Ã—7^{3}) = 2Ã—2Ã—3Ã—5Ã—7 = 420

**(x)** 157464

By using prime factorization method

âˆ›157464 = âˆ› (2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3) = âˆ› (2^{3}Ã—3^{3}Ã—3^{3}Ã—3^{3}) = 2Ã—3Ã—3Ã—3 = 54

**(xi)** 1157625

By using prime factorization method

âˆ›1157625 = âˆ› (3Ã—3Ã—3Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7) = âˆ› (3^{3}Ã—5^{3}Ã—7^{3}) = 3Ã—5Ã—7 = 105

**(xii)** 33698267

By using prime factorization method

âˆ›33698267 = âˆ› (17Ã—17Ã—17Ã—19Ã—19Ã—19) = âˆ› (17^{3}Ã—19^{3}) = 17Ã—19 = 323

**5. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.**

**Solution:**

Firstly letâ€™s find the prime factors for 3600

3600 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5

= 2^{3}Â Ã— 3^{2}Â Ã— 5^{2}Â Ã— 2

Since only one triples is formed and three factors remained ungrouped in triples.

The given number 3600 is not a perfect cube.

To make it a perfect cube we have to multiply it by (2 Ã— 2 Ã— 3 Ã— 5) = 60

3600 Ã— 60 = 216000

Cube root of 216000 is

âˆ›216000 = âˆ› (60Ã—60Ã—60) = âˆ› (60^{3}) = 60

âˆ´ the smallest number which when multiplied with 3600 will make the product a perfect cube is 60 and the cube root of the product is 60.

**6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.**

**Solution:**

The prime factors of 210125 are

210125 = 5 Ã— 5 Ã— 5 Ã— 41 Ã— 41

Since, one triples remained incomplete, 210125 is not a perfect cube.

To make it a perfect cube we need to multiply the factors by 41, we will get 2 triples as 23Â and 41^{3}.

And the product become:

210125 Ã— 41 = 8615125

8615125 = 5 Ã— 5 Ã— 5 Ã— 41 Ã— 41 Ã— 41

Cube root of product = âˆ›8615125 = âˆ› (5Ã—41) = 205

**7. What is the smallest number by which 8192 must be divided so that quotient is a perfect cube? Also, find the cube root of the quotient so obtained.**

**Solution**:

The prime factors of 8192 are

8192 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2 = 2^{3}Ã—2^{3}Ã—2^{3}Ã—2

Since, one triples remain incomplete, hence 8192 is not a perfect cube.

So, we divide 8192 by 2 to make its quotient a perfect cube.

8192/2 = 4096

4096 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2 = 2^{3}Ã—2^{3}Ã—2^{3}Ã—2^{3}

Cube root of 4096 = âˆ›4096 = âˆ› (2^{3}Ã—2^{3}Ã—2^{3}Ã—2^{3}) = 2Ã—2Ã—2Ã—2 = 16

**8. Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.**

**Solution:**

Let us consider the ratio 1:2:3 as x, 2x and 3x

According to the question,

X^{3}Â + (2x)^{ 3}Â + (3x)^{ 3}Â = 98784

x^{3}Â + 8x^{3}Â + 27x^{3}Â = 98784

36x^{3}Â = 98784

x^{3} = 98784/36

= 2744

x = âˆ›2744 = âˆ› (2Ã—2Ã—2Ã—7Ã—7Ã—7) = 2Ã—7 = 14

So, the numbers are,

x = 14

2x = 2Â Ã— 14 = 28

3x = 3Â Ã— 14 = 42

9. **The volume of a cube is 9261000 m ^{3}. Find the side of the cube.**

Given, volume of cube = 9261000 m^{3}

Let us consider the side of cube be â€˜aâ€™ metre

So, a^{3} = 9261000

a = âˆ›9261000 = âˆ› (2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7) = âˆ› (2^{3}Ã—3^{3}Ã—5^{3}Ã—7^{3}) = 2Ã—3Ã—5Ã—7 = 210

âˆ´ the side of cube = 210 metre

## EXERCISE 4.4 PAGE NO: 4.30

**1. Find the cube roots of each of the following integers:
(i)-125 (ii) -5832
(iii)-2744000 (iv) -753571
(v) -32768**

**Solution:**

**(i)** -125

The cube root of -125 is

-125 = âˆ›-125 = -âˆ›125 = âˆ› (5Ã—5Ã—5) = -5

**(ii)** -5832

The cube root of -5832 is

-5832 = âˆ›-5832 = -âˆ›5832

To find the cube root of 5832, we shall use the method of unit digits.

Let us consider the number 5832. Where, unit digit of 5832 = 2

Unit digit in the cube root of 5832 will be 8

After striking out the units, tens and hundreds digits of 5832,

Now we left with 5 only.

We know that 1 is the Largest number whose cube is less than or equal to 5.

So, the tens digit of the cube root of 5832 is 1.

âˆ›-5832 = -âˆ›5832 = -18

**(iii)** -2744000

âˆ›-2744000 = -âˆ›2744000

We shall use the method of factorization to find the cube root of 2744000

So letâ€™s find the prime factors for 2744000

2744000 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7

Now by grouping the factors into triples of equal factors, we get,

2744000 = (2Ã—2Ã—2) Ã— (2Ã—2Ã—2) Ã— (5Ã—5Ã—5) Ã— (7Ã—7Ã—7)

Since all the prime factors of 2744000 is grouped in to triples of equal factors and no factor is left over.

So now take one factor from each group and by multiplying we get,

2Ã—2Ã—5Ã—7 = 140

Thereby we can say that 2744000 is a cube of 140

âˆ´ âˆ›-2744000 = -âˆ›2744000 = -140

**(iv)** -753571

âˆ›-753571 = -âˆ›753571

We shall use the unit digit method,

Let us consider the number 753571, where unit digit = 1

Unit digit in the cube root of 753571 will be 1

After striking out the units, tens and hundreds digits of 753571,

Now we left with 753.

We know that 9 is the Largest number whose cube is less than or equal to 753(9^{3}<753<10^{3}).

So, the tens digit of the cube root of 753571 is 9.

âˆ›753571 = 91

âˆ›-753571 = -âˆ›753571 = -91

**(v)** -32768

âˆ›-32765 = -âˆ›32768

We shall use the unit digit method,

Let us consider the Number = 32768, where unit digit = 8

Unit digit in the cube root of 32768 will be 2

After striking out the units, tens and hundreds digits of 32768,

Now we left with 32.

As we know that 9 is the Largest number whose cube is less than or equals to 32(3^{3}<32<4^{3}).

So, the tens digit of the cube root of 32768 is 3.

âˆ›32765 = 32

âˆ›-32765 = -âˆ›32768 = -32

**2. Show that:**

**(i) âˆ›27 Ã— âˆ›64 = âˆ› (27Ã—64)**

**(ii) âˆ› (64Ã—729) = âˆ›64 Ã— âˆ›729**

**(iii) âˆ› (-125Ã—216) = âˆ›-125 Ã— âˆ›216**

**(iv) âˆ› (-125Ã—-1000) = âˆ›-125 Ã— âˆ›-1000**

**Solution:**

**(i) **âˆ›27 Ã— âˆ›64 = âˆ› (27Ã—64)

Let us consider LHS âˆ›27 Ã— âˆ›64

âˆ›27 Ã— âˆ›64 = âˆ›(3Ã—3Ã—3) Ã— âˆ›(4Ã—4Ã—4)

= 3Ã—4

= 12

Let us consider RHS âˆ› (27Ã—64)

âˆ› (27Ã—64) = âˆ› (3Ã—3Ã—3Ã—4Ã—4Ã—4)

= 3Ã—4

= 12

âˆ´ LHS = RHS, the given equation is verified.

**(ii) **âˆ› (64Ã—729) = âˆ›64 Ã— âˆ›729

Let us consider LHS âˆ› (64Ã—729)

âˆ› (64Ã—729) = âˆ› (4Ã—4Ã—4Ã—9Ã—9Ã—9)

= 4Ã—9

= 36

Let us consider RHS âˆ›64 Ã— âˆ›729

âˆ›64 Ã— âˆ›729 = âˆ›(4Ã—4Ã—4) Ã— âˆ›(9Ã—9Ã—9)

= 4Ã—9

= 36

âˆ´ LHS = RHS, the given equation is verified.

**(iii) **âˆ› (-125Ã—216) = âˆ›-125 Ã— âˆ›216

Let us consider LHS âˆ› (-125Ã—216)

âˆ› (-125Ã—216) = âˆ› (-5Ã—-5Ã—-5Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3)

= -5Ã—2Ã—3

= -30

Let us consider RHS âˆ›-125 Ã— âˆ›216

âˆ›-125 Ã— âˆ›216 = âˆ›(-5Ã—-5Ã—-5) Ã— âˆ›(2Ã—2Ã—2Ã—3Ã—3Ã—3)

= -5Ã—2Ã—3

= -30

âˆ´ LHS = RHS, the given equation is verified.

**(iv) **âˆ› (-125Ã—-1000) = âˆ›-125 Ã— âˆ›-1000

Let us consider LHS âˆ› (-125Ã—-1000)

âˆ› (-125Ã—-1000) = âˆ› (-5Ã—-5Ã—-5Ã—-10Ã—-10Ã—-10)

= -5Ã—-10

= 50

Let us consider RHS âˆ›-125 Ã— âˆ›-1000

âˆ›-125 Ã— âˆ›-1000 = âˆ›(-5Ã—-5Ã—-5) Ã— âˆ›(-10Ã—-10Ã—-10)

= -5Ã—-10

= 50

âˆ´ LHS = RHS, the given equation is verified.

**3. Find the cube root of each of the following numbers:
(i) 8Ã—125Â Â **

**(ii)Â -1728Ã—216
(iii)Â -27Ã—2744Â **

**(iv)Â -729Ã—-15625**

**Solution:**

**(i) **8Ã—125**Â Â **

We know that for any two integers a and b,Â âˆ› (aÃ—b) = âˆ›a Ã— âˆ›b

By using the property

âˆ› (8Ã—125) = âˆ›8 Ã— âˆ›125

= âˆ› (2Ã—2Ã—2) Ã— âˆ› (5Ã—5Ã—5)

= 2Ã—5

= 10

**(ii)Â **-1728Ã—216

We know that for any two integers a and b,Â âˆ› (aÃ—b) = âˆ›a Ã— âˆ›b

By using the property

âˆ› (-1728Ã—216) = âˆ›-1728 Ã— âˆ›216

We shall use the unit digit method

Let us consider the number 1728, where Unit digit = 8

The unit digit in the cube root of 1728 will be 2

After striking out the units, tens and hundreds digits of the given number, we are left with the 1.

We know 1 is the largest number whose cube is less than or equal to 1.

So, the tens digit of the cube root of 1728 = 1

âˆ›1728 = 12

Now, letâ€™s find the prime factors for 216 = 2Ã—2Ã—2Ã—3Ã—3Ã—3

By grouping the factors in triples of equal factor, we get,

216 = (2Ã—2Ã—2) Ã— (3Ã—3Ã—3)

By taking one factor from each group we get,

âˆ›216 = 2Ã—3 = 6

âˆ´ by equating the values in the given equation we get,

âˆ› (-1728Ã—216) = âˆ›-1728 Ã— âˆ›216

= -12 Ã— 6

= -72

**(iii)Â **-27Ã—2744**Â **

We know that for any two integers a and b,Â âˆ› (aÃ—b) = âˆ›a Ã— âˆ›b

By using the property

âˆ› (-27Ã—2744) = âˆ›-27 Ã— âˆ›2744

We shall use the unit digit method

Let us consider the number 2744, where Unit digit = 4

The unit digit in the cube root of 2744 will be 4

After striking out the units, tens and hundreds digits of the given number, we are left with the 2.

We know 2 is the largest number whose cube is less than or equal to 2.

So, the tens digit of the cube root of 2744 = 2

âˆ›2744 = 14

Now, letâ€™s find the prime factors for 27 = 3Ã—3Ã—3

By grouping the factors in triples of equal factor, we get,

27 = (3Ã—3Ã—3)

Cube root of 27 is

âˆ›27 = 3

âˆ´ by equating the values in the given equation we get,

âˆ› (-27Ã—2744) = âˆ›-27 Ã— âˆ›2744

= -3 Ã— 14

= -42

**(iv)Â **-729Ã—-15625

We know that for any two integers a and b,Â âˆ› (aÃ—b) = âˆ›a Ã— âˆ›b

By using the property

âˆ› (-729Ã—-15625) = âˆ›-729 Ã— âˆ›-15625

We shall use the unit digit method

Let us consider the number 15625, where Unit digit = 5

The unit digit in the cube root of 15625 will be 5

After striking out the units, tens and hundreds digits of the given number, we are left with the 15.

We know 15 is the largest number whose cube is less than or equal to 15(2^{3}<15<3^{3}).

So, the tens digit of the cube root of 15625 = 2

âˆ›15625 = 25

Now, letâ€™s find the prime factors for 729 = 9Ã—9Ã—9

By grouping the factors in triples of equal factor, we get,

729 = (9Ã—9Ã—9)

Cube root of 729 is

âˆ›729 = 9

âˆ´ by equating the values in the given equation we get,

âˆ› (-729Ã—-15625) = âˆ›-729 Ã— âˆ›-15625

= -9 Ã— -25

= 225

**4. Evaluate:**

**(i) âˆ› (4 ^{3} Ã— 6^{3})**

**(ii) âˆ› (8Ã—17Ã—17Ã—17)**

**(iii) âˆ› (700Ã—2Ã—49Ã—5)**

**(iv) 125 âˆ›a ^{6} – âˆ›125a^{6}**

**Solution:**

**(i) **âˆ› (4^{3} Ã— 6^{3})

We know that for any two integers a and b,Â âˆ› (aÃ—b) = âˆ›a Ã— âˆ›b

By using the property

âˆ› (4^{3} Ã— 6^{3}) = âˆ›4^{3} Ã— âˆ›6^{3}

= 4 Ã— 6

= 24

**(ii) **âˆ› (8Ã—17Ã—17Ã—17)

We know that for any two integers a and b,Â âˆ› (aÃ—b) = âˆ›a Ã— âˆ›b

By using the property

âˆ› (8Ã—17Ã—17Ã—17) = âˆ›8 Ã— âˆ›17Ã—17Ã—17

= âˆ›2^{3} Ã— âˆ›17^{3}

= 2 Ã— 17

= 34

**(iii) **âˆ› (700Ã—2Ã—49Ã—5)

Firstly let us find the prime factors for the above numbers

âˆ› (700Ã—2Ã—49Ã—5) = âˆ› (2Ã—2Ã—5Ã—5Ã—7Ã—2Ã—7Ã—7Ã—5)

= âˆ› (2^{3}Ã—5^{3}Ã—7^{3})

= 2Ã—5Ã—7

= 70

**(iv) **125 âˆ›a^{6} – âˆ›125a^{6}

125 âˆ›a^{6} – âˆ›125a^{6} = 125 âˆ›(a^{2})^{3} – âˆ›5^{3}(a^{2})^{3}

= 125a^{2} â€“ 5a^{2}

= 120a^{2}

**5. Find the cube root of each of the following rational numbers:**

**(i) -125/729**

**(ii) 10648/12167**

**(iii) -19683/24389**

**(iv) 686/-3456**

**(v) -39304/-42875**

**Solution:**

**(i) **-125/729

Let us find the prime factors of 125 and 729

-125/729 = – (âˆ› (5Ã—5Ã—5)) / (âˆ› (9Ã—9Ã—9))

= – (âˆ› (5^{3})) / (âˆ› (9^{3}))

= – 5/9

**(ii) **10648/12167

Let us find the prime factors of 10648 and 12167

10648/12167 = (âˆ› (2Ã—2Ã—2Ã—11Ã—11Ã—11)) / (âˆ› (23Ã—23Ã—23))

= (âˆ› (2^{3}Ã—11^{3})) / (âˆ› (23^{3}))

= (2Ã—11)/23

= 22/23

**(iii) **-19683/24389

Let us find the prime factors of 19683 and 24389

-19683/24389 = -(âˆ› (3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3)) / (âˆ› (29Ã—29Ã—29))

= – (âˆ› (3^{3}Ã—3^{3}Ã—3^{3})) / (âˆ› (29^{3}))

= – (3Ã—3Ã—3)/29

= – 27/29

**(iv) **686/-3456

Let us find the prime factors of 686 and -3456

686/-3456 = = -(âˆ› (2Ã—7Ã—7Ã—7)) / (âˆ› (2^{7}Ã—2^{3}))

= – (âˆ› (2Ã—7^{3})) / (âˆ› (2^{7}Ã—2^{3}))

= – (âˆ› (7^{3})) / (âˆ› (2^{6}Ã—2^{3}))

= – 7/(2Ã—2Ã—2)

= – 7/8

**(v) **-39304/-42875

Let us find the prime factors of -39304 and -42875

-39304/-42875 = -(âˆ› (2Ã—2Ã—2Ã—17Ã—17Ã—17)) / -(âˆ› (5Ã—5Ã—5Ã—7Ã—7Ã—7))

= – (âˆ› (2^{3}Ã—17^{3})) / -(âˆ› (5^{3}Ã—7^{3}))

= – (2Ã—17)/-( 5Ã—7)

= – 34/-35

= 34/35

**6. Find the cube root of each of the following rational numbers:
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331**

**Solution:**

**(i) **0.001728

0.001728 = 1728/1000000

âˆ› (0.001728) = âˆ›1728 / âˆ›1000000

Let us find the prime factors of 1728 and 1000000

âˆ›(0.001728) = âˆ›(2^{3}Ã—2^{3}Ã—3^{3}) / âˆ›(100^{3})

= (2Ã—2Ã—3)/100

= 12/100

= 0.12

**(ii) **0.003375

0.003375 = 3375/1000000

âˆ› (0.003375) = âˆ›3375 / âˆ›1000000

Let us find the prime factors of 3375 and 1000000

âˆ›(0.003375) = âˆ›(3^{3}Ã—5^{3}) / âˆ›(100^{3})

= (3Ã—5)/100

= 15/100

= 0.15

**(iii) **0.001

0.001 = 1/1000

âˆ› (0.001) = âˆ›1 / âˆ›1000

= 1/ âˆ›10^{3}

= 1/10

= 0.1

**(iv) **1.331

1.331 = 1331/1000

âˆ› (1.331) = âˆ›1331 / âˆ›1000

Let us find the prime factors of 1331 and 1000

âˆ›(1.331) = âˆ›(11^{3}) / âˆ›(10^{3})

= 11/10

= 1.1

**7. Evaluate each of the following:**

**(i) âˆ›27 + âˆ›0.008 + âˆ›0.064**

**(ii) âˆ›1000 + âˆ›0.008 – âˆ›0.125**

**(iii) âˆ›(729/216) Ã— 6/9**

**(iv) âˆ›(0.027/0.008) Ã· âˆ›(0.09/0.04) – 1
(v) âˆ›(0.1Ã—0.1Ã—0.1Ã—13Ã—13Ã—13)**

**Solution:**

**(i) **âˆ›27 + âˆ›0.008 + âˆ›0.064

Let us simplify

âˆ› (3Ã—3Ã—3) + âˆ› (0.2Ã—0.2Ã—0.2) + âˆ› (0.4Ã—0.4Ã—0.4)

âˆ› (3)^{3} + âˆ› (0.2)^{3} + âˆ› (0.4)^{3}

3 + 0.2 + 0.4

3.6

**(ii) **âˆ›1000 + âˆ›0.008 – âˆ›0.125

Let us simplify

âˆ› (10Ã—10Ã—10) + âˆ› (0.2Ã—0.2Ã—0.2) – âˆ› (0.5Ã—0.5Ã—0.5)

âˆ› (10)^{3} + âˆ› (0.2)^{3} – âˆ› (0.5)^{3}

10 + 0.2 – 0.5

9.7

**(iii) **âˆ› (729/216) Ã— 6/9

Let us simplify

âˆ› (9Ã—9Ã—9/6Ã—6Ã—6) Ã— 6/9

(âˆ› (9)^{3} / âˆ› (6)^{3})Ã— 6/9

9/6 Ã— 6/9

1

**(iv)** âˆ›(0.027/0.008) Ã· âˆš(0.09/0.04) – 1

Let us simplify âˆ›(0.027/0.008) Ã· âˆš (0.09/0.04)

âˆ›(0.3Ã—0.3Ã—0.3/0.2Ã—0.2Ã—0.2) Ã· âˆš(0.3Ã—0.3/0.2Ã—0.2)

(âˆ›(0.3)^{3} / âˆ›(0.2)^{3}) Ã· (âˆš(0.3)^{2} / âˆš(0.2)^{2})

(0.3/0.2) Ã· (0.3/0.2) -1

(0.3/0.2 Ã— 0.2/0.3) -1

1 â€“ 1

0

**(v) **âˆ›(0.1Ã—0.1Ã—0.1Ã—13Ã—13Ã—13)

âˆ›(0.1^{3}Ã—13^{3})

0.1 Ã— 13 = 1.3

**8. Show that:**

**(i) âˆ› (729)/ âˆ› (1000) = âˆ› (729/1000)**

**(ii) âˆ› (-512)/ âˆ› (343) = âˆ› (-512/343)**

**Solution:**

**(i) **âˆ› (729)/ âˆ› (1000) = âˆ› (729/1000)

Let us consider LHS âˆ› (729)/ âˆ› (1000)

âˆ› (729)/ âˆ› (1000) = âˆ› (9Ã—9Ã—9)/ âˆ› (10Ã—10Ã—10)

** = **âˆ› (9^{3}/10^{3})

= 9/10

Let us consider RHS âˆ› (729/1000)

âˆ› (729/1000) = âˆ› (9Ã—9Ã—9/10Ã—10Ã—10)

** = **âˆ› (9^{3}/10^{3})

= 9/10

âˆ´ LHS = RHS

**(ii) **âˆ› (-512)/ âˆ› (343) = âˆ› (-512/343)

Let us consider LHS âˆ› (-512)/ âˆ› (343)

âˆ› (-512)/ âˆ› (343) = âˆ›-(8Ã—8Ã—8)/ âˆ› (7Ã—7Ã—7)

** = **âˆ›-(8^{3}/7^{3})

= -8/7

Let us consider RHS âˆ› (-512/343)

âˆ› (-512/343) = = âˆ›-(8Ã—8Ã—8/7Ã—7Ã—7)

** = **âˆ›-(8^{3}/7^{3})

= -8/7

âˆ´ LHS = RHS

**9. Fill in the blanks:**

**(i) âˆ›(125Ã—27) = 3 Ã— â€¦**

**(ii) âˆ›(8Ã—â€¦) = 8
(iii) âˆ›1728 = 4 Ã— â€¦**

**(iv) âˆ›480 = âˆ›3Ã—2Ã— âˆ›..**

**(v) âˆ›â€¦ = âˆ›7 Ã— âˆ›8**

**(vi) âˆ›..= âˆ›4 Ã— âˆ›5 Ã— âˆ›6**

**(vii) âˆ›(27/125) = â€¦/5**

**(viii) âˆ›(729/1331) = 9/â€¦**

**(ix) âˆ›(512/â€¦) = 8/13**

**Solution:**

**(i) **âˆ›(125Ã—27) = 3 Ã— â€¦

Let us consider LHS âˆ›(125Ã—27)

âˆ›(125Ã—27) = âˆ›(5Ã—5Ã—5Ã—3Ã—3Ã—3)

= âˆ›(5^{3}Ã—3^{3})

= 5Ã—3 or 3Ã—5

**(ii) **âˆ›(8Ã—â€¦) = 8

Let us consider LHS âˆ›(8Ã—â€¦)

âˆ›(8Ã—8Ã—8) = âˆ›8^{3} = 8

**(iii) **âˆ›1728 = 4 Ã— â€¦

Let us consider LHS

âˆ›1728 = âˆ›(2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3)

= âˆ›(2^{3}Ã—2^{3}Ã—3^{3})

= 2Ã—2Ã—3

= 4Ã—3

**(iv) **âˆ›480 = âˆ›3Ã—2Ã— âˆ›..

Let us consider LHS

âˆ›480 = âˆ›(2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—5)

= âˆ›(2^{3}Ã—2^{2}Ã—3Ã—5)

= âˆ›2^{3}Ã— âˆ›3 Ã— âˆ›2Ã—2Ã—5

= 2 Ã— âˆ›3 Ã— âˆ›20

**(v) **âˆ›â€¦ = âˆ›7 Ã— âˆ›8

Let us consider RHS

âˆ›7 Ã— âˆ›8 = âˆ›(7 Ã— 8)

= âˆ›56

**(vi) **âˆ›..= âˆ›4 Ã— âˆ›5 Ã— âˆ›6

Let us consider RHS

âˆ›4 Ã— âˆ›5 Ã— âˆ›6 = âˆ›(4 Ã— 5 Ã— 6)

= âˆ›120

**(vii) **âˆ›(27/125) = â€¦/5

Let us consider LHS

âˆ›(27/125) = âˆ›(3Ã—3Ã—3)/(5Ã—5Ã—5)

= âˆ›(3^{3})/(5^{3})

= 3/5

**(viii) **âˆ›(729/1331) = 9/â€¦

Let us consider LHS

âˆ›(729/1331) = âˆ›(9Ã—9Ã—9)/(11Ã—11Ã—11)

= âˆ›(9^{3})/(11^{3})

= 9/11

**(ix) **âˆ›(512/â€¦) = 8/13

Let us consider LHS

âˆ›(512/â€¦) = âˆ›(2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2)

= âˆ›(2^{3}Ã—2^{3}Ã—2^{3})

= 2Ã—2Ã—2

= 8

So, 8/âˆ›â€¦ = 8/13

when numerators are same the denominators will also become equal.

8 Ã— 13 = 8 Ã— âˆ›â€¦

âˆ›â€¦= 13

â€¦ = (13)^{3}

= 2197

**10. The volume of a cubical box is 474. 552 cubic metres. Find the length of each side of the box.**

**Solution:**

Volume of a cubical box is 474.552 cubic metres

V = 8^{3},

Let â€˜Sâ€™ be the side of the cube

8^{3}Â = 474.552 cubic metres

8 = âˆ›474.552

= âˆ› (474552/1000)

Let us factorise 474552 into prime factors, we get:

474552 = 2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—13Ã—13Ã—13

By grouping the factors in triples of equal factors, we get:

474552 = (2Ã—2Ã—2) Ã— (3Ã—3Ã—3) Ã— (13Ã—13Ã—13)

Now, âˆ›474.552 = âˆ› ((2Ã—2Ã—2) Ã— (3Ã—3Ã—3) Ã— (13Ã—13Ã—13))

= 2Ã—3Ã—13

= 78

Also,

âˆ›1000 = âˆ› (10Ã—10Ã—10)

= âˆ› (10)^{3}

= 10

So now let us equate in the above equation we get,

8 = âˆ› (474552/1000)

= 78/10

= 7.8

âˆ´ length of the side is 7.8m.

**11. Three numbers are to one another 2:3:4. The sum of their cubes is 0.334125. Find the numbers.**

**Solution:**

Let us consider the ratio 2:3:4 be 2a, 3a, and 4a.

So according to the question:

(2a)^{3}+ (3a)^{3}+(4a)^{3}Â = 0.334125

8a^{3}+27 a^{3}+64 a^{3}Â = 0.334125

99a^{3}Â = 0.334125

a^{3} = 334125/1000000Ã—99

= 3375/1000000

a = âˆ› (3375/1000000)

= âˆ›((15Ã—15Ã—15)/ 100Ã—100Ã—100)

= 15/100

= 0.15

âˆ´ The numbers are:

2Ã—0.15 = 0.30

3Ã—0.15 = 0.45

4Ã—0.15 = 0.6

**12. Find the side of a cube whose volume is 24389/216m ^{3}.**

**Solution:**

Volume of the side s =Â 24389/216 = v

V= 8^{.3}

8 = âˆ›v

= âˆ›(24389/216)

By performing factorisation we get,

= âˆ›(29Ã—29Ã—29/2Ã—2Ã—2Ã—3Ã—3Ã—3)

= 29/(2Ã—3)

= 29/6

âˆ´ The length of the side is 29/6.

**13. Evaluate:**

**(i) âˆ›36 Ã— âˆ›384**

**(ii) âˆ›96 Ã— âˆ›144**

**(iii) âˆ›100 Ã— âˆ›270**

**(iv) âˆ›121 Ã— âˆ›297**

**Solution:**

**(i) **âˆ›36 Ã— âˆ›384

We know that âˆ›a Ã— âˆ›b = âˆ› (aÃ—b)

By using the above formula let us simplify

âˆ›36 Ã— âˆ›384 = âˆ› (36Ã—384)

The prime factors of 36 and 384 are

= âˆ› (2Ã—2Ã—3Ã—3) Ã— (2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3)

= âˆ›(2^{3}Ã—2^{3}Ã—2^{3}Ã—3^{3})

= 2Ã—2Ã—2Ã—3

= 24

**(ii) **âˆ›96 Ã— âˆ›144

We know that âˆ›a Ã— âˆ›b = âˆ› (aÃ—b)

By using the above formula let us simplify

âˆ›96 Ã— âˆ›144 = âˆ› (96Ã—144)

The prime factors of 96 and 144 are

= âˆ› (2Ã—2Ã—2Ã—2Ã—2Ã—3) Ã— (2Ã—2Ã—2Ã—2Ã—3Ã—3)

= âˆ›(2^{3}Ã—2^{3}Ã—2^{3}Ã—3^{3})

= 2Ã—2Ã—2Ã—3

= 24

**(iii) **âˆ›100 Ã— âˆ›270

We know that âˆ›a Ã— âˆ›b = âˆ› (aÃ—b)

By using the above formula let us simplify

âˆ›100 Ã— âˆ›270 = âˆ› (100Ã—270)

The prime factors of 100 and 270 are

= âˆ› (2Ã—2Ã—5Ã—5) Ã— (2Ã—3Ã—3Ã—3Ã—5)

= âˆ›(2^{3}Ã—3^{3}Ã—5^{3})

= 2Ã—3Ã—5

= 30

**(iv) **âˆ›121 Ã— âˆ›297

We know that âˆ›a Ã— âˆ›b = âˆ› (aÃ—b)

By using the above formula let us simplify

âˆ›121 Ã— âˆ›297 = âˆ› (121Ã—297)

The prime factors of 121 and 297 are

= âˆ› (11Ã—11) Ã— (3Ã—3Ã—3Ã—11)

= âˆ›(11^{3}Ã—3^{3})

= 11Ã—3

= 33

**14. Find the cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that
(i)Â 3048625 = 3375 Ã— 729
(ii)Â 20346417 = 9261 Ã— 2197
(iii)Â 210644875 = 42875 Ã— 4913
(iv)Â 57066625 = 166375 Ã— 343**

**Solution:**

**(i)Â **3048625 = 3375 Ã— 729

By taking the cube root for the whole we get,

âˆ›3048625 = âˆ›3375 Ã— âˆ›729

Now perform factorization

= âˆ›3Ã—3Ã—3Ã—5Ã—5Ã—5 Ã— âˆ›9Ã—9Ã—9

= âˆ›3^{3}Ã—5^{3} Ã— âˆ›9^{3}

= 3Ã—5Ã—9

= 135

**(ii)**Â 20346417 = 9261 Ã— 2197

By taking the cube root for the whole we get,

âˆ›20346417 = âˆ›9261 Ã— âˆ›2197

Now perform factorization

= âˆ›3Ã—3Ã—3Ã—7Ã—7Ã—7 Ã— âˆ›13Ã—13Ã—13

= âˆ›3^{3}Ã—7^{3} Ã— âˆ›13^{3}

= 3Ã—7Ã—13

= 273

**(iii)Â **210644875 = 42875 Ã— 4913

By taking the cube root for the whole we get,

âˆ›210644875 = âˆ›42875 Ã— âˆ›4913

Now perform factorization

= âˆ›5Ã—5Ã—5Ã—7Ã—7Ã—7 Ã— âˆ›17Ã—17Ã—17

= âˆ›5^{3}Ã—7^{3} Ã— âˆ›17^{3}

= 5Ã—7Ã—17

= 595

**(iv)Â **57066625 = 166375 Ã— 343

By taking the cube root for the whole we get,

âˆ›57066625 = âˆ›166375 Ã— âˆ›343

Now perform factorization

= âˆ›5Ã—5Ã—5Ã—11Ã—11Ã—11 Ã— âˆ›7Ã—7Ã—7

= âˆ›5^{3}Ã—11^{3} Ã— âˆ›7^{3}

= 5Ã—11Ã—7

= 385

**15. Find the unit of the cube **

**root of the following numbers:
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616**

**Solution:**

(i) 226981

The given number is 226981.

Unit digit of 226981 = 1

The unit digit of the cube root of 226981 = 1

(ii) 13824

The given number is 13824.

Unit digit of 13824 = 4

The unit digit of the cube root of 13824 = 4

(iii) 571787

The given number is 571787.

Unit digit of 571787 = 7

The unit digit of the cube root of 571787 = 7

(iv) 175616

The given number is 175616.

Unit digit of 175616 = 6

The unit digit of the cube root of 175616 = 6

**16. Find the tens digit of the cube root of each of the numbers in Q.No.15.
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616**

**Solution:**

**(i)** 226981

The given number is 226981.

Unit digit of 226981 = 1

The unit digit in the cube root of 226981 = 1

After striking out the units, tens and hundreds digits of 226981, now we left with 226 only.

We know that 6 is the Largest number whose cube root is less than or equal to 226(6^{3}<226<7^{3}).

âˆ´ The tens digit of the cube root of 226981 is 6.

**(ii)** 13824

The given number is 13824.

Unit digit of 13824 = 4

The unit digit in the cube root of 13824 = 4

After striking out the units, tens and hundreds digits of 13824, now we left with 13 only.

We know that 2 is the Largest number whose cube root is less than or equal to 13(2^{3}<13<3^{3}).

âˆ´ The tens digit of the cube root of 13824 is 2.

**(iii)** 571787

The given number is 571787.

Unit digit of 571787 = 7

The unit digit in the cube root of 571787 = 3

After striking out the units, tens and hundreds digits of 571787, now we left with 571 only.

We know that 8 is the Largest number whose cube root is less than or equals to 571(8^{3}<571<9^{3}).

âˆ´ The tens digit of the cube root of 571787 is 8.

**(iv)** 175616

The given number is 175616.

Unit digit of 175616 = 6

The unit digit in the cube root of 175616 = 6

After striking out the units, tens and hundreds digits of 175616, now we left with 175 only.

We know that 5 is the Largest number whose cube root is less than or equals to 175(5^{3}<175<6^{3}).

âˆ´ The tens digit of the cube root of 175616 is 5.

## EXERCISE 4.5 PAGE NO: 4.36

**Making use of the cube root table, find the cube root of the following (correct to three decimal places):**

**1. 7**

**Solution:**

As we know that 7 lies between 1 and 100 so by using cube root table we get,

âˆ›7 = 1.913

âˆ´ the answer is 1.913

**2. 70**

**Solution:**

As we know that 70 lies between 1 and 100 so by using cube root table from column x

We get,

âˆ›70 = 4.121

âˆ´ the answer is 4.121

**3. 700**

**Solution:**

700 = 70Ã—10

By using cube root table 700 will be in the column âˆ›10xÂ against 70.

So we get,

âˆ›700 = 8.879

âˆ´ the answer is 8.879

**4. 7000**

**Solution:**

7000 = 70Ã—100

âˆ›7000 = âˆ›(7Ã—1000) = âˆ›7 Ã— âˆ›1000

By using cube root table,

We get,

âˆ›7 = 1.913

âˆ›1000 = 10

âˆ›7000 = âˆ›7 Ã— âˆ›1000

= 1.913 Ã— 10

= 19.13

âˆ´ the answer is 19.13

**5. 1100**

**Solution:**

1100 = 11Ã—100

âˆ›1100 = âˆ›(11Ã—100) = âˆ›11 Ã— âˆ›100

By using cube root table,

We get,

âˆ›11 = 2.224

âˆ›100 = 4.6642

âˆ›1100 = âˆ›11 Ã— âˆ›100

= 2.224 Ã— 4.642

= 10.323

âˆ´ the answer is 10.323

**6.780**

**Solution:**

780 = 78Ã—10

By using cube root table 780 would be in column âˆ›10xÂ against 78.

We get,

âˆ›780 = 9.205

**7. 7800**

**Solution:**

7800 = 78Ã—100

âˆ›7800 = âˆ›(78Ã—100) = âˆ›78 Ã— âˆ›100

By using cube root table,

We get,

âˆ›78 = 4.273

âˆ›100 = 4.6642

âˆ›7800 = âˆ›78 Ã— âˆ›100

= 4.273 Ã— 4.642

= 19.835

âˆ´ the answer is 19.835

**8. 1346**

**Solution:**

Let us find the factors by using factorisation method,

We get,

1346 = 2Ã—673

âˆ›1346 = âˆ›(2Ã—676) = âˆ›2 Ã— âˆ›673

Since, 670<673<680 = âˆ›670 < âˆ›673 < âˆ›680

By using cube root table,

âˆ›670 = 8.750

âˆ›680 = 8.794

For the difference (680-670) which is 10.

So the difference in the values = 8.794 – 8.750 = 0.044

For the difference (673-670) which is 3.

So the difference in the values = (0.044/10) Ã— 3 = 0.0132

âˆ›673 = 8.750 + 0.013 = 8.763

âˆ›1346 = âˆ›2 Ã— âˆ›673

= 1.260 Ã— 8.763

= 11.041

âˆ´ the answer is 11.041

**9. 250**

**Solution:**

250 = 25Ã—100

By using cube root table 250 would be in column âˆ›10xÂ against 25.

We get,

âˆ›250 = 6.3

âˆ´ the answer is 6.3

**10. 5112**

**Solution:**

Let us find the factors by using factorisation method,

âˆ›5112 = âˆ›2Ã—2Ã—2Ã—3Ã—3Ã—71

= âˆ›2^{3}Ã—3^{2}Ã—71

= 2 Ã— âˆ›3^{2} Ã— âˆ›71

= 2 Ã— âˆ›9 Ã— âˆ›71

From cube root table we get,

âˆ›9 = 2.080

âˆ›71 = 4.141

âˆ›5112 = 2 Ã— âˆ›9 Ã— âˆ›71

= 2 Ã— 2.080 Ã— 4.141

= 17.227

âˆ´ the answer is 17.227

**11. 9800**

**Solution:**

âˆ›9800 = âˆ›98 Ã— âˆ›100

From cube root table we get,

âˆ›98 = 4.610

âˆ›100 = 4.642

âˆ›9800 = âˆ›98 Ã— âˆ›100

= 4.610 Ã— 4.642

= 21.40

âˆ´ the answer is 21.40

**12. 732**

**Solution:**

âˆ›732

We know that value of âˆ›732 will lie between âˆ›730 and âˆ›740

From cube root table we get,

âˆ›730 = 9.004

âˆ›740 = 9.045

By using unitary method,

Difference between the values (740 â€“ 730 = 10)

So, the difference in cube root values will be = 9.045 â€“ 9.004 = 0.041

Difference between the values (732 â€“ 730 = 2)

So, the difference in cube root values will be = (0.041/10) Ã—2 = 0.008

âˆ›732 = 9.004+0.008 = 9.012

âˆ´ the answer is 9.012

**13. 7342**

**Solution:**

âˆ›7342

We know that value of âˆ›7342 will lie between âˆ›7300 and âˆ›7400

From cube root table we get,

âˆ›7300 = 19.39

âˆ›7400 = 19.48

By using unitary method,

Difference between the values (7400 â€“ 7300 = 100)

So, the difference in cube root values will be = 19.48 â€“ 19.39 = 0.09

Difference between the values (7342 â€“ 7300 = 42)

So, the difference in cube root values will be = (0.09/100) Ã— 42 = 0.037

âˆ›7342 = 19.39+0.037 = 19.427

âˆ´ the answer is 19.427

**14. 133100**

**Solution:**

âˆ›133100 = âˆ› (1331Ã—100)

= âˆ›1331 Ã— âˆ›100

= âˆ› 11^{3} Ã— âˆ›100

= 11 Ã— âˆ›100

From cube root table we get,

âˆ›100 = 4.462

âˆ›133100 = 11 Ã— âˆ›100

= 11 Ã— 4.462

= 51.062

âˆ´ the answer is 51.062

**15. 37800**

**Solution:**

âˆ›37800

Firstly let us find the factors for 37800

âˆ›37800 = âˆ›(2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—175)

= âˆ›(2^{3}Ã—3^{3}Ã—175)

= 6 Ã— âˆ›175

We know that value of âˆ›175 will lie between âˆ›170 and âˆ›180

From cube root table we get,

âˆ›170 = 5.540

âˆ›180 = 5.646

By using unitary method,

Difference between the values (180 â€“ 170 = 10)

So, the difference in cube root values will be = 5.646 â€“ 5.540 = 0.106

Difference between the values (175 â€“ 170 = 5)

So, the difference in cube root values will be = (0.106/10) Ã— 5 = 0.053

âˆ›175 = 5.540 + 0.053 = 5.593

âˆ›37800 = 6 Ã— âˆ›175

= 6 Ã— 5.593

= 33.558

âˆ´ the answer is 33.558

**16. 0.27**

**Solution:**

âˆ›0.27 = âˆ›(27/100) = âˆ›27/âˆ›100

From cube root table we get,

âˆ›27 = 3

âˆ›100 = 4.642

âˆ›0.27 = âˆ›27/âˆ›100

= 3/4.642

= 0.646

âˆ´ the answer is 0.646

**17. 8.6**

**Solution:**

âˆ›8.6 = âˆ›(86/10) = âˆ›86/âˆ›10

From cube root table we get,

âˆ›86 = 4.414

âˆ›10 = 2.154

âˆ›8.6 = âˆ›86/âˆ›10

= 4.414/2.154

= 2.049

âˆ´ the answer is 2.049

**18. 0.86**

**Solution: **

âˆ›0.86 = âˆ›(86/100) = âˆ›86/âˆ›100

From cube root table we get,

âˆ›86 = 4.414

âˆ›100 = 4.642

âˆ›8.6 = âˆ›86/âˆ›100

= 4.414/4.642

= 0.9508

âˆ´ the answer is 0.951

**19. 8.65**

**Solution:**

âˆ›8.65 = âˆ›(865/100) = âˆ›865/âˆ›100

We know that value of âˆ›865 will lie between âˆ›860 and âˆ›870

From cube root table we get,

âˆ›860 = 9.510

âˆ›870 = 9.546

âˆ›100 = 4.642

By using unitary method,

Difference between the values (870 â€“ 860 = 10)

So, the difference in cube root values will be = 9.546 â€“ 9.510 = 0.036

Difference between the values (865 â€“ 860 = 5)

So, the difference in cube root values will be = (0.036/10) Ã— 5 = 0.018

âˆ›865 = 9.510 + 0.018 = 9.528

âˆ›8.65 = âˆ›865/âˆ›100

= 9.528/4.642

= 2.0525

âˆ´ the answer is 2.053

**20. 7532**

**Solution:**

âˆ›7532

We know that value of âˆ›7532 will lie between âˆ›7500 and âˆ›7600

From cube root table we get,

âˆ›7500 = 19.57

âˆ›7600 = 19.66

By using unitary method,

Difference between the values (7600 â€“ 7500 = 100)

So, the difference in cube root values will be = 19.66 â€“ 19.57 = 0.09

Difference between the values (7532 â€“ 7500 = 32)

So, the difference in cube root values will be = (0.09/100) Ã— 32 = 0.029

âˆ›7532 = 19.57 + 0.029 = 19.599

âˆ´ the answer is 19.599

**21. 833**

**Solution:**

âˆ›833

We know that value of âˆ›833 will lie between âˆ›830 and âˆ›840

From cube root table we get,

âˆ›830 = 9.398

âˆ›840 = 9.435

By using unitary method,

Difference between the values (840 â€“ 830 = 10)

So, the difference in cube root values will be = 9.435 â€“ 9.398 = 0.037

Difference between the values (833 â€“ 830 = 3)

So, the difference in cube root values will be = (0.037/10) Ã—3 = 0.011

âˆ›833 = 9.398+0.011 = 9.409

âˆ´ the answer is 9.409

**22. 34.2**

**Solution:**

âˆ›34.2 = âˆ›(342/10) = âˆ›342/âˆ›10

We know that value of âˆ›342 will lie between âˆ›340 and âˆ›350

From cube root table we get,

âˆ›340 = 6.980

âˆ›350 = 7.047

âˆ›10 = 2.154

By using unitary method,

Difference between the values (350 â€“ 340 = 10)

So, the difference in cube root values will be = 7.047 â€“ 6.980 = 0.067

Difference between the values (342 â€“ 340 = 2)

So, the difference in cube root values will be = (0.067/10) Ã— 2 = 0.013

âˆ›342 = 6.980 + 0.013 = 6.993

âˆ›34.2 = âˆ›342/âˆ›10

= 6.993/2.154

= 3.246

âˆ´ the answer is 3.246

**23. What is the length of the side of a cube whose volume is 275 cm ^{3}. Make use of the table for the cube root.**

**Solution:**

The given volume of the cube = 275cm^{3}

Let us consider the side of the cube as â€˜aâ€™cm

a^{3} = 275

a = âˆ›275

We know that value of âˆ›275 will lie between âˆ›270 and âˆ›280

From cube root table we get,

âˆ›270 = 6.463

âˆ›280 = 6.542

By using unitary method,

Difference between the values (280 â€“ 270 = 10)

So, the difference in cube root values will be = 6.542 â€“ 6.463 = 0.079

Difference between the values (275 â€“ 270 = 5)

So, the difference in cube root values will be = (0.079/10) Ã— 5 = 0.0395

âˆ›275 = 6.463 + 0.0395 = 6.5025

âˆ´ the answer is 6.503cm