**RD Sharma Solutions for Class 8 Maths Chapter 4 – Cubes and Cube Roots** are provided here. RD Sharma Solutions help students secure good marks in the annual examination, as they provide extensive knowledge about the subject. BYJU’S expert team has solved the questions from the RD Sharma Class 8 textbook in a step-by-step format in detail, which will help students strengthen their knowledge of the concepts covered in this chapter.

Chapter 4 – Cubes and Cube Roots contains five exercises, and the RD Sharma Class 8 Solutions available on this page provide 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

## RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots

### Access Answers to RD Sharma Solutions for Class 8 Maths Chapter 4 – Cubes and Cube Roots

## 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 = 7× 7 × 7 = 343

**(ii) **12

The cube of 12 is

12 = 12× 12× 12 = 1728

**(iii) **16

Cube of 16 is

16 = 16× 16× 16 = 4096

**(iv) **21

Cube of 21 is

21 = 21 × 21 × 21 = 9261

**(v) **40

The cube of 40 is

40 = 40× 40× 40 = 64000

**(vi)** 55

Cube of 55 is

55 = 55× 55× 55 = 166375

**(vii) **100

The cube of 100 is

100 = 100× 100× 100 = 1000000

**(viii) **302

The cube of 302 is

302 = 302× 302× 302 = 27543608

**(ix) **301

The cube of 301 is

301 = 301× 301× 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} = 1 × 1 × 1 = 1

2^{3} = 2 × 2 × 2 = 8

3^{3} = 3 × 3 × 3 = 27

4^{3} = 4 × 4 × 4 = 64

5^{3} = 5 × 5 × 5 = 125

6^{3} = 6 × 6 × 6 = 216

7^{3} = 7 × 7 × 7 = 343

8^{3} = 8 × 8 × 8 = 512

9^{3} = 9 × 9 × 9 = 729

10^{3} = 10 × 10 × 10 = 1000

∴ 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} +…+ 9^{3} by the above pattern.

**Solution:**

According to the given pattern,

1^{3} + 2^{3} + 3^{3} +…+ 9^{3}

1^{3} + 2^{3} + 3^{3} +…+ n^{3} = (1+2+3+…+n)^{ 2}

So when n = 10

1^{3} + 2^{3} + 3^{3} +…+ 9^{3 }+ 10^{3} = (1+2+3+…+10)^{ 2}

= (55)^{2} = 55×55 = 3025

**4. Write the cubes of 5 natural numbers, which are multiples of 3 and verify the followings:
“The cube of a natural number which is a multiple of 3 is a multiple of 27’**

**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} = 3 × 3 × 3 = 27

6^{3} = 6 × 6 × 6 = 216

9^{3} = 9 × 9 × 9 = 729

12^{3} = 12 × 12 × 12 = 1728

15^{3} = 15 × 15 × 15 = 3375

We found that all the cubes are divisible by 27

∴ “The cube of a natural number which is a multiple of 3 is a multiple of 27’

**5. Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, …) and verify the following:
“The 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 1’**

**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} = 4 × 4 × 4 = 64

7^{3} = 7 × 7 × 7 = 343

10^{3} = 10 × 10 × 10 = 1000

13^{3} = 13 × 13 × 13 = 2197

16^{3} = 16 × 16 × 16 = 4096

We found that all these cubes, when divided by ‘3’, leave the remainder 1.

∴ the statement “The 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 1’ is true.

**6. Write the cubes 5 natural numbers of the form 3n+2(i.e.5,8,11….) and verify the following:
“The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is divided by 3, the remainder is 2’**

**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} = 5 × 5 × 5 = 125

8^{3} = 8 × 8 × 8 = 512

11^{3} = 11 × 11 × 11 = 1331

14^{3} = 14 × 14 × 14 = 2744

17^{3} = 17 × 17 × 17 = 4913

We found that all these cubes, when divided by ‘3’, leave the remainder 2.

∴ the statement “The cube of a natural number of the form 3n+2 is a natural number of the same form, i.e. when it is divided by 3, the remainder is 2’ is true.

**7. Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
“The 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} = 7 × 7 × 7 = 343

14^{3} = 14 × 14 × 14 = 2744

21^{3} = 21× 21× 21 = 9261

28^{3} = 28 × 28 × 28 = 21952

35^{3} = 35 × 35 × 35 = 42875

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

∴ The statement “The 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 × 2 × 2 × 2 × 2 × 2 = 2^{6} = (2^{2})^{3} = 4^{3}

Hence, it’s a perfect cube.

**(ii)** 216

First, find thefactors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2^{3} × 3^{3} = 6^{3}

Hence, it’s a perfect cube.

**(iii)** 243

First, find thefactors of 243

243 = 3 × 3 × 3 × 3 × 3 = 3^{5} = 3^{3} × 3^{2}

Hence, it’s not a perfect cube.

**(iv)** 1000

First, find thefactors of 1000

1000 = 2 × 2 × 2 × 5 × 5 × 5 = 2^{3} × 5^{3} = 10^{3}

Hence, it’s a perfect cube.

**(v)** 1728

First, find thefactors of 1728

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2^{6} × 3^{3} = (4 × 3 )^{3} = 12^{3}

Hence, it’s a perfect cube.

**(vi)** 3087

First, find thefactors of 3087

3087 = 3 × 3 × 7 × 7 × 7 = 3^{2} × 7^{3}

Hence, it’s not a perfect cube.

**(vii)** 4608

First, find thefactors of 4608

4608 = 2 × 2 × 3 × 113

Hence, it’s not a perfect cube.

**(viii)** 106480

First, find thefactors of 106480

106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11

Hence, it’s not a perfect cube.

**(ix)** 166375

First, find thefactors of 166375

166375= 5 × 5 × 5 × 11 × 11 × 11 = 5^{3} × 11^{3} = 55^{3}

Hence, it’s a perfect cube.

**(x)** 456533

First, find thefactors of 456533

456533= 11 × 11 × 11 × 7 × 7 × 7 = 11^{3} × 7^{3} = 77^{3}

Hence, it’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} × 3^{3} = 6^{3}

It’s a cube of even natural number.

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

It’s a cube of even natural number.

**(iii)** 729 = 3^{3} × 3^{3} = 9^{3}

It’s not a cube of even natural number.

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

It’s a cube of even natural number.

**(v)** 3375 = 3^{3} × 5^{3} = 15^{3}

It’s not a cube of even natural number.

**(vi)** 13824 = 2^{9 }× 3^{3} = (2^{3})^{3} × 3^{3} = 8^{3}×3^{3 }= 24^{3}

It’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 × 5 × 5 × 5 = 5^{3}

It’s a cube of odd natural number.

**(ii)** 343 = 7 × 7 × 7 = 7^{3}

It’s a cube of odd natural number.

**(iii)** 1728 = 2^{6} × 3^{3} = 4^{3} × 3^{3} = 12^{3}

It’s not a cube of odd natural number. As 12 is an even number.

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

It’s not a cube of odd natural number. As 64 is an even number.

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

It’s not a cube of odd natural number. As 32 is an even number.

**(vi)** 6859 = 19 × 19 × 19 = 19^{3}

It’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 × 3 × 3 × 5 × 5

= 3^{3} × 5^{2}

∴To make a perfect cube, we need to multiply the product by 5.

**(ii)** 1323

First, find the factors of 1323

1323 = 3 × 3 × 3 × 7 × 7

= 3^{3} × 7^{2}

∴To make a perfect cube, we need to multiply the product by 7.

**(iii)** 2560

First, find the factors of 2560

2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

= 2^{3} × 2^{3} × 2^{3} × 5

∴To make a perfect cube, we need to multiply the product by 5 × 5 = 25.

**(iv)** 7803

First, find the factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3^{3} × 17^{2}

∴To make a perfect cube, we need to multiply the product by 17.

**(v)** 107811

First, find the factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 3^{3} × 3 × 11^{3}

∴To make a perfect cube, we need to multiply the product by 3 × 3 = 9.

**(vi)** 35721

First, find the factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 3^{3} × 3^{3} × 7^{2}

∴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 × 3 × 3 × 5 × 5

= 3^{3} × 5^{2}

Since 675 is not a perfect cube.

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

∴ 25 is the required smallest number.

**(ii)** 8640

First, find the prime factors of 8640

8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

= 2^{3} × 2^{3} × 3^{3} × 5

Since 8640 is not a perfect cube.

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

∴ is the required smallest number.

**(iii)** 1600

First, find the prime factors of 1600

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 2^{3} × 2^{3} × 5^{2}

Since 1600 is not a perfect cube.

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

∴ 25 is the required smallest number.

**(iv)** 8788

First, find the prime factors of 8788

8788 = 2 × 2 × 13 × 13 × 13

= 2^{2} × 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 the quotient, and we know that 2197 is a perfect cube

∴ 4 is the required smallest number.

**(v)** 7803

First, find the prime factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3^{3} × 17^{2}

Since 7803 is not a perfect cube.

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

∴ 289 is the required smallest number.

**(vi)** 107811

First, find the prime factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 3^{3} × 11^{3} × 3

Since 107811 is not a perfect cube.

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

∴ 3 is the required smallest number.

**(vii)** 35721

First, find the prime factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 3^{3} × 3^{3} × 7^{2}

Since 35721 is not a perfect cube.

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

∴ 49 is the required smallest number.

**(viii)** 243000

First, find the prime factors of 243000

243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5

= 2^{3} × 3^{3} × 5^{3} × 3^{2}

Since 243000 is not a perfect cube.

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

∴ 9 is the required smallest number.

**13. Prove that if a number is trebled then its cube is 27 times 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 × a = 3a

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

∴ The 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} = 27a^{3}

Hence, the number will become 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} = 64a^{3}

Hence, the number will become 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} = 125a^{3}

Hence, the number will become 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 a cube = 64 m^{2}

Let the length of the edge of cube be ‘a’ metre

a^{2} = 64

a = √ 64

= 8m

Now, the volume of cube = a^{3}

a^{3} = 8^{3} = 8 × 8 × 8

= 512m^{3}

∴ The 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 the cube = 384 m^{2}

Let us consider the length of each edge of the cube be ‘a’ metre

6a^{2} = 384

a^{2} = 384/6

= 64

a = √64

= 8m

Now, the volume of cube = a^{3}

a^{3} = 8^{3} = 8 × 8 × 8

= 512m^{3}

∴ The volume of a cube is 512m^{3}

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

**Solution:**

(i) {(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) {(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 the 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} = 1

∴ The unit digit of the cube of 31 is always 1

**109**

To find the 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} = 729

∴ The unit digit of the cube of 109 is always 9

**388**

To find the 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} = 512

∴ The unit digit of the cube of 388 is always 2

**4276**

To find the 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} = 216

∴ The unit digit of the cube of 4276 is always 6

**5922**

To find the 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} = 8

∴ The unit digit of the cube of 5922 is always 8

**77774**

To find the 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} = 64

∴ The unit digit of the cube of 77774 is always 4

**44447**

To find the 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} = 343

∴ The unit digit of the cube of 44447 is always 3

**125125125**

To find the 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} = 125

∴ The unit digit of the 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
3×a |
Column III
3×a×b |
Column IV
b |

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

+15 | +23 | +12 | 125 |

42 | 158 | 237 | |

42 | 8 | 7 | 5 |

∴ The cube of 35 is 42875

**(ii) **56

We have, a = 5 and b = 6

Column I
a |
Column II
3×a |
Column III
3×a×b |
Column IV
b |

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

+50 | +56 | +21 | 126 |

175 | 506 | 561 | |

175 | 6 | 1 | 6 |

∴ The cube of 56 is 175616

**(iii) 72**

We have, a = 7 and b = 2

Column I
a |
Column II
3×a |
Column III
3×a×b |
Column IV
b |

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

+30 | +8 | +0 | 8 |

373 | 302 | 84 | |

373 | 2 | 4 | 8 |

∴ 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 × 2 × 2 × 2 × 2 × 2

= 2^{3} × 2^{3}

= 4^{3}

Hence, it’s a perfect cube.

(ii) 216

Firstly let us find the prime factors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3

= 2^{3} × 3^{3}

= 6^{3}

Hence, it’s a perfect cube.

(iii) 243

Firstly let us find the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3

= 3^{3} × 3^{2}

Hence, it’s not a perfect cube.

(iv) 1728

Firstly let us find the prime factors of 1728

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 2^{3} × 2^{3} × 3^{3}

= 12^{3}

Hence, it’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:**

The only non-perfect cube in the 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 × 3 × 3 × 3 × 3 = 3^{3} × 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 × 3 × 3 × 3 × 3 = 3^{3} × 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} is also even.
(ii) If n is odd, then n^{3} is also odd.
(ii) If n leaves the remainder 1 when divided by 3, then n^{3} also leaves 1 as the remainder when divided by 3.
(iv) If a natural number n is of the form 3p+2, then n^{3} is also a number of the same type.**

**Solution:**

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

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

So now, cubes of 2, 4 and 6 are

2^{3} = 8

4^{3} = 64

6^{3} = 216

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

Statement is verified.

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

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

So now, cubes of 3, 5 and 7 are

3^{3} = 27

5^{3} = 125

7^{3} = 343

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

Statement is verified.

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

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

So now, the cube of 4, 7, and 10 are

4^{3} = 64

7^{3} = 343

10^{3} = 1000

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

Hence, the statement is verified.

**(iv)** If a natural number n is of the form 3p+2, then n^{3} is 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, the cube of 5, 8 and 10 are

5^{3} = 125

8^{3} = 512

11^{3} = 1331

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

125 = 3 × 41 + 2

512 = 3 × 170 + 2

1331 = 3 × 443 + 2

Hence, the 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}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}divides b

^{3}. (viii) If a

^{2}ends in 9, then a

^{3}ends in 7. (ix) If a

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

^{3}ends in 25. (x) If a

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

^{3}ends in an odd number of zeros.

**Solution:**

(i) 392 is a perfect cube.

Firstly let’s find the prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 2^{3} × 7^{2}

Hence the statement is False.

(ii) 8640 is not a perfect cube.

Prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2^{3} × 2^{3} × 3^{3} × 5

Hence the statement is True

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

Statement is True.

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

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

We know 64 is a perfect cube = 4 × 4 × 4, and it ends with 4.

Hence the statement is False.

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

Statement is False.

Because in the 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 the 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} 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} ends in 9, then a^{3} ends in 7.

Statement is False.

Let a = 7

7^{2} = 49 and 7^{3} = 343

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

Statement is False.

Since, when a = 20

a^{2} = 20^{2} = 400 and a^{3} = 8000 (a^{3} doesn’t end with 25)

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

Statement is False.

Since, when a = 100

a^{2} = 100^{2} = 10000 and a^{3} = 100^{3} = 1000000 (a^{3} doesn’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} = -11× -11× -11 = -1331

(ii) -12

The cube of 12 is

(-12)^{3} = -12× -12× -12 = -1728

(iii) -21

The cube of 21 is

(-21)^{3} = -21× -21× -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 × 2 × 2 × 2 × 2 × 2

= 2^{3} × 2^{3}

= 4^{3}

∴ 64 is a perfect cube of negative integer – 4.

**(ii)** -1056

The prime factors of 1056 are

1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11

1056 is not a perfect cube.

∴ -1056 is not a cube of a negative integer.

**(iii)** -2197

The prime factors of 2197 are

2197 = 13 × 13 × 13

= 13^{3}

∴ 2197 is a perfect cube of negative integer – 13.

**(iv)** -2744

The prime factors of 2744 are

2744 = 2 × 2 × 2 × 7 × 7 × 7

= 2^{3} × 7^{3}

= 14^{3}

2744 is a perfect cube.

∴ -2744 is a cube of negative integer – 14.

**(v)** -42875

The prime factors of 42875 are

42875 = 5 × 5 × 5 × 7 × 7 × 7

= 5^{3} × 7^{3}

= 35^{3}

42875 is a perfect cube.

∴ -42875 is a cube of negative integer – 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 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= 2^{3} × 3^{3} × 3^{3}

= 18^{3}

5832 is a perfect cube.

∴ -5832 is a cube of negative integer – 18.

**(ii)** -2744000

The prime factors of 2744000 are

2744000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7

= 2^{3} × 2^{3}× 5^{3} × 7^{3}

2744000 is a perfect cube.

∴ -2744000 is a cube of negative integer – 140.

**4. Find the cube of:
(i) 7/9 (ii) -8/11
(iii) 12/7 (iv) -13/8
(v) **

**\(\begin{array}{l}2\frac{2}{5}\end{array} \)**

**(vi)\(\begin{array}{l}3\frac{1}{4}\end{array} \)
(vii) 0.3 (viii) 1.5**

(ix) 0.08 (x) 2.1

(ix) 0.08 (x) 2.1

**Solution:**

**(i)** 7/9

The cube of 7/9 is

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

**(ii)** -8/11

The cube of -8/11 is

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

**(iii)** 12/7

The cube of 12/7 is

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

**(iv)** -13/8

The cube of -13/8 is

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

**(v)**

The cube of 12/5 is

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

**(vi)**

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} = 0.3×0.3×0.3 = 0.027

**(viii)** 1.5

The cube of 1.5 is

(1.5)^{3} = 1.5×1.5×1.5 = 3.375

**(ix)** 0.08

The cube of 0.08 is

(0.08)^{3} = 0.08×0.08×0.08 = 0.000512

**(x)** 2.1

The cube of 2.1 is

(2.1)^{3} = 2.1×2.1×2.1 = 9.261

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

**Solution:**

**(i)** 27/64

We have,

27/64 = (3×3×3)/ (4×4×4) = 3^{3}/4^{3} = (3/4)^{3}

∴ 27/64 is a cube of 3/4.

**(ii)** 125/128

We have,

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

∴ 125/128 is not a perfect cube.

**(iii)** 0.001331

We have,

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

∴ 0.001331 is a perfect cube of 11/100

**(iv)** 0.04

We have,

4/10 = (2×2)/(2×5) = 2^{2}/(2×5)

∴ 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, …
(i) 64
(ii) 512
(iii) 1728**

**Solution:**

**(i)** 64

Let’s perform subtraction

64 – 1 = 63

63 – 7 = 56

56 – 19 =37

37 – 37 = 0

Subtraction is performed 4 times.

∴ The cube root of 64 is 4.

**(ii)** 512

Let’s perform subtraction

512 – 1 = 511

511 – 7 = 504

504 – 19 = 485

485 – 37 = 448

448 – 61 = 387

387 – 91 = 296

296 – 127 = 169

169 – 169 = 0

Subtraction is performed 8 times.

∴ The cube root of 512 is 8.

**(iii)** 1728

Let’s perform subtraction

1728 – 1 = 1727

1727 – 7 = 1720

1720 – 19 = 1701

1701 – 37 = 1664

1664 – 91 = 1512

1512 – 127 = 1385

1385 – 169 = 1216

1216 – 217 = 999

999 – 271 = 728

728 – 331 = 397

397 – 397 = 0

Subtraction is performed 12 times.

∴ The 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

Let’s perform subtraction

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

The next number to be subtracted is 91, which is greater than 5

∴130 is not a perfect cube.

**(ii)** 345

Let’s perform subtraction

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

The next number to be subtracted is 169, which is greater than 2

∴ 345 is not a perfect cube

**(iii)** 792

Let’s perform subtraction

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

The next number to be subtracted is 271, which is greater than 63

∴ 792 is not a perfect cube

**(iv)** 1331

Let’s perform subtraction

1331 – 1 = 1330

1330 – 7 = 1323

1323 – 19 = 1304

1304 – 37 = 1267

1267 – 61 = 1206

1206 – 91 = 1115

1115 – 127 = 988

988 – 169 = 819

819 – 217 = 602

602 – 271 = 331

331 – 331 = 0

Subtraction is performed 11 times

The cube root of 1331 is 11

∴ 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 the previous question, there are three numbers which are not perfect cubes.

**(i)** 130

Let’s perform subtraction

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

The next number to be subtracted is 91, which is greater than 5

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

130 – 5 = 125 (which is a perfect cube of 5)

**(ii)** 345

Let’s perform subtraction

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

The next number to be subtracted is 169, which is greater than 2

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

345 – 2 = 343 (which is a perfect cube of 7)

**(iii)** 792

Let’s perform subtraction

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

The next number to be subtracted is 271, which is greater than 63

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

792 – 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 the prime factorisation method

∛343 = ∛ (7×7×7) = 7

**(ii)** 2744

By using the prime factorisation method

∛2744 = ∛ (2×2×2×7×7×7) = ∛ (2^{3}×7^{3}) = 2×7 = 14

**(iii)** 4913

By using the prime factorisation method,

∛4913 = ∛ (17×17×17) = 17

**(iv)** 1728

By using the prime factorisation method,

∛1728 = ∛(2×2×2×2×2×2×3×3×3) = ∛ (2^{3}×2^{3}×3^{3}) = 2×2×3 = 12

**(v)** 35937

By using the prime factorisation method,

∛35937 = ∛ (3×3×3×11×11×11) = ∛ (3^{3}×11^{3}) = 3×11 = 33

**(vi)** 17576

By using the prime factorisation method,

∛17576 = ∛ (2×2×2×13×13×13) = ∛ (2^{3}×13^{3}) = 2×13 = 26

**(vii)** 134217728

By using the prime factorisation method

∛134217728 = ∛ (2^{27}) = 2^{9} = 512

**(viii)** 48228544

By using the prime factorisation method

∛48228544 = ∛ (2×2×2×2×2×2×7×7×7×13×13×13) = ∛ (2^{3}×2^{3}×7^{3}×13^{3}) = 2×2×7×13 = 364

**(ix)** 74088000

By using the prime factorisation method

∛74088000 = ∛ (2×2×2×2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (2^{3}×2^{3}×3^{3}×5^{3}×7^{3}) = 2×2×3×5×7 = 420

**(x)** 157464

By using the prime factorisation method

∛157464 = ∛ (2×2×2×3×3×3×3×3×3×3×3×3) = ∛ (2^{3}×3^{3}×3^{3}×3^{3}) = 2×3×3×3 = 54

**(xi)** 1157625

By using the prime factorisation method

∛1157625 = ∛ (3×3×3×5×5×5×7×7×7) = ∛ (3^{3}×5^{3}×7^{3}) = 3×5×7 = 105

**(xii)** 33698267

By using the prime factorisation method

∛33698267 = ∛ (17×17×17×19×19×19) = ∛ (17^{3}×19^{3}) = 17×19 = 323

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

**Solution:**

Firstly let’s find the prime factors for 3600

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 2^{3} × 3^{2} × 5^{2} × 2

Since only one triple is formed and three factors remain 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 × 2 × 3 × 5) = 60

3600 × 60 = 216000

The cube root of 216000 is

∛216000 = ∛ (60×60×60) = ∛ (60^{3}) = 60

∴ the smallest number, which, when multiplied by 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 × 5 × 5 × 41 × 41

Since one triple 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 23 and 41^{3}.

And the product becomes:

210125 × 41 = 8615125

8615125 = 5 × 5 × 5 × 41 × 41 × 41

Cube root of product = ∛8615125 = ∛ (5×41) = 205

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

**Solution**:

The prime factors of 8192 are

8192 = 2×2×2×2×2×2×2×2×2×2×2 = 2^{3}×2^{3}×2^{3}×2

Since one triple remains 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 = 2×2×2×2×2×2×2×2×2×2×2×2 = 2^{3}×2^{3}×2^{3}×2^{3}

Cube root of 4096 = ∛4096 = ∛ (2^{3}×2^{3}×2^{3}×2^{3}) = 2×2×2×2 = 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} + (2x)^{ 3} + (3x)^{ 3} = 98784

x^{3} + 8x^{3} + 27x^{3} = 98784

36x^{3} = 98784

x^{3} = 98784/36

= 2744

x = ∛2744 = ∛ (2×2×2×7×7×7) = 2×7 = 14

So, the numbers are,

x = 14

2x = 2 × 14 = 28

3x = 3 × 14 = 42

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

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

Let us consider the side of the cube to be ‘a’ metre

So, a^{3} = 9261000

a = ∛9261000 = ∛ (2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (2^{3}×3^{3}×5^{3}×7^{3}) = 2×3×5×7 = 210

∴ 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 = ∛-125 = -∛125 = -∛ (5×5×5) = -5

**(ii)** -5832

The cube root of -5832 is

-5832 = ∛-5832 = -∛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 are 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.

∛-5832 = -∛5832 = -18

**(iii)** -2744000

∛-2744000 = -∛2744000

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

So let’s find the prime factors for 2744000

2744000 = 2×2×2×2×2×2×5×5×5×7×7×7

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

2744000 = (2×2×2) × (2×2×2) × (5×5×5) × (7×7×7)

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

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

2×2×5×7 = 140

Thereby we can say that 2744000 is a cube of 140

∴ ∛-2744000 = -∛2744000 = -140

**(iv)** -753571

∛-753571 = -∛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 are 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.

∛753571 = 91

∛-753571 = -∛753571 = -91

**(v)** -32768

∛-32765 = -∛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 are left with 32.

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

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

∛32765 = 32

∛-32765 = -∛32768 = -32

**2. Show that:**

**(i) ∛27 × ∛64 = ∛ (27×64)**

**(ii) ∛ (64×729) = ∛64 × ∛729**

**(iii) ∛ (-125×216) = ∛-125 × ∛216**

**(iv) ∛ (-125×-1000) = ∛-125 × ∛-1000**

**Solution:**

**(i) **∛27 × ∛64 = ∛ (27×64)

Let us consider LHS ∛27 × ∛64

∛27 × ∛64 = ∛(3×3×3) × ∛(4×4×4)

= 3×4

= 12

Let us consider RHS ∛ (27×64)

∛ (27×64) = ∛ (3×3×3×4×4×4)

= 3×4

= 12

∴ LHS = RHS, the given equation is verified.

**(ii) **∛ (64×729) = ∛64 × ∛729

Let us consider LHS ∛ (64×729)

∛ (64×729) = ∛ (4×4×4×9×9×9)

= 4×9

= 36

Let us consider RHS ∛64 × ∛729

∛64 × ∛729 = ∛(4×4×4) × ∛(9×9×9)

= 4×9

= 36

∴ LHS = RHS, the given equation is verified.

**(iii) **∛ (-125×216) = ∛-125 × ∛216

Let us consider LHS ∛ (-125×216)

∛ (-125×216) = ∛ (-5×-5×-5×2×2×2×3×3×3)

= -5×2×3

= -30

Let us consider RHS ∛-125 × ∛216

∛-125 × ∛216 = ∛(-5×-5×-5) × ∛(2×2×2×3×3×3)

= -5×2×3

= -30

∴ LHS = RHS, the given equation is verified.

**(iv) **∛ (-125×-1000) = ∛-125 × ∛-1000

Let us consider LHS ∛ (-125×-1000)

∛ (-125×-1000) = ∛ (-5×-5×-5×-10×-10×-10)

= -5×-10

= 50

Let us consider RHS ∛-125 × ∛-1000

∛-125 × ∛-1000 = ∛(-5×-5×-5) × ∛(-10×-10×-10)

= -5×-10

= 50

∴ LHS = RHS, the given equation is verified.

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

**(ii) -1728×216
(iii) -27×2744 **

**(iv) -729×-15625**

**Solution:**

**(i) **8×125** **

We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b

By using the property

∛ (8×125) = ∛8 × ∛125

= ∛ (2×2×2) × ∛ (5×5×5)

= 2×5

= 10

**(ii) **-1728×216

We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b

By using the property

∛ (-1728×216) = ∛-1728 × ∛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

∛1728 = 12

Now, let’s find the prime factors for 216 = 2×2×2×3×3×3

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

216 = (2×2×2) × (3×3×3)

By taking one factor from each group, we get,

∛216 = 2×3 = 6

∴ by equating the values in the given equation we get,

∛ (-1728×216) = ∛-1728 × ∛216

= -12 × 6

= -72

**(iii) **-27×2744** **

We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b

By using the property

∛ (-27×2744) = ∛-27 × ∛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

∛2744 = 14

Now, let’s find the prime factors for 27 = 3×3×3

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

27 = (3×3×3)

The cube root of 27 is

∛27 = 3

∴ by equating the values in the given equation we get,

∛ (-27×2744) = ∛-27 × ∛2744

= -3 × 14

= -42

**(iv) **-729×-15625

We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b

By using the property

∛ (-729×-15625) = ∛-729 × ∛-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

∛15625 = 25

Now, let’s find the prime factors for 729 = 9×9×9

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

729 = (9×9×9)

The cube root of 729 is

∛729 = 9

∴ by equating the values in the given equation we get,

∛ (-729×-15625) = ∛-729 × ∛-15625

= -9 × -25

= 225

**4. Evaluate:**

**(i) ∛ (4 ^{3} × 6^{3})**

**(ii) ∛ (8×17×17×17)**

**(iii) ∛ (700×2×49×5)**

**(iv) 125 ∛a ^{6} – ∛125a^{6}**

**Solution:**

**(i) **∛ (4^{3} × 6^{3})

We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b

By using the property

∛ (4^{3} × 6^{3}) = ∛4^{3} × ∛6^{3}

= 4 × 6

= 24

**(ii) **∛ (8×17×17×17)

We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b

By using the property

∛ (8×17×17×17) = ∛8 × ∛17×17×17

= ∛2^{3} × ∛17^{3}

= 2 × 17

= 34

**(iii) **∛ (700×2×49×5)

Firstly let us find the prime factors for the above numbers

∛ (700×2×49×5) = ∛ (2×2×5×5×7×2×7×7×5)

= ∛ (2^{3}×5^{3}×7^{3})

= 2×5×7

= 70

**(iv) **125 ∛a^{6} – ∛125a^{6}

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

= 125a^{2} – 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 = – (∛ (5×5×5)) / (∛ (9×9×9))

= – (∛ (5^{3})) / (∛ (9^{3}))

= – 5/9

**(ii) **10648/12167

Let us find the prime factors of 10648 and 12167

10648/12167 = (∛ (2×2×2×11×11×11)) / (∛ (23×23×23))

= (∛ (2^{3}×11^{3})) / (∛ (23^{3}))

= (2×11)/23

= 22/23

**(iii) **-19683/24389

Let us find the prime factors of 19683 and 24389

-19683/24389 = -(∛ (3×3×3×3×3×3×3×3×3)) / (∛ (29×29×29))

= – (∛ (3^{3}×3^{3}×3^{3})) / (∛ (29^{3}))

= – (3×3×3)/29

= – 27/29

**(iv) **686/-3456

Let us find the prime factors of 686 and -3456

686/-3456 = = -(∛ (2×7×7×7)) / (∛ (2^{7}×3^{3}))

= – (∛ (2×7^{3})) / (∛ (2^{7}×3^{3}))

= – (∛ (7^{3})) / (∛ (2^{6}×3^{3}))

= – 7/(2×2×3)

= – 7/12

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

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

-39304/-42875 = -(∛ (2×2×2×17×17×17)) / -(∛ (5×5×5×7×7×7))

= – (∛ (2^{3}×17^{3})) / -(∛ (5^{3}×7^{3}))

= – (2×17)/-( 5×7)

= – 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

∛ (0.001728) = ∛1728 / ∛1000000

Let us find the prime factors of 1728 and 1000000

∛(0.001728) = ∛(2^{3}×2^{3}×3^{3}) / ∛(100^{3})

= (2×2×3)/100

= 12/100

= 0.12

**(ii) **0.003375

0.003375 = 3375/1000000

∛ (0.003375) = ∛3375 / ∛1000000

Let us find the prime factors of 3375 and 1000000

∛(0.003375) = ∛(3^{3}×5^{3}) / ∛(100^{3})

= (3×5)/100

= 15/100

= 0.15

**(iii) **0.001

0.001 = 1/1000

∛ (0.001) = ∛1 / ∛1000

= 1/ ∛10^{3}

= 1/10

= 0.1

**(iv) **1.331

1.331 = 1331/1000

∛ (1.331) = ∛1331 / ∛1000

Let us find the prime factors of 1331 and 1000

∛(1.331) = ∛(11^{3}) / ∛(10^{3})

= 11/10

= 1.1

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

**(i) ∛27 + ∛0.008 + ∛0.064**

**(ii) ∛1000 + ∛0.008 – ∛0.125**

**(iii) ∛(729/216) × 6/9**

**(iv) ∛(0.027/0.008) ÷ ∛(0.09/0.04) – 1
(v) ∛(0.1×0.1×0.1×13×13×13)**

**Solution:**

**(i) **∛27 + ∛0.008 + ∛0.064

Let us simplify

∛ (3×3×3) + ∛ (0.2×0.2×0.2) + ∛ (0.4×0.4×0.4)

∛ (3)^{3} + ∛ (0.2)^{3} + ∛ (0.4)^{3}

3 + 0.2 + 0.4

3.6

**(ii) **∛1000 + ∛0.008 – ∛0.125

Let us simplify

∛ (10×10×10) + ∛ (0.2×0.2×0.2) – ∛ (0.5×0.5×0.5)

∛ (10)^{3} + ∛ (0.2)^{3} – ∛ (0.5)^{3}

10 + 0.2 – 0.5

9.7

**(iii) **∛ (729/216) × 6/9

Let us simplify

∛ (9×9×9/6×6×6) × 6/9

(∛ (9)^{3} / ∛ (6)^{3})× 6/9

9/6 × 6/9

1

**(iv)** ∛(0.027/0.008) ÷ √(0.09/0.04) – 1

Let us simplify ∛(0.027/0.008) ÷ √ (0.09/0.04)

∛(0.3×0.3×0.3/0.2×0.2×0.2) ÷ √(0.3×0.3/0.2×0.2)

(∛(0.3)^{3} / ∛(0.2)^{3}) ÷ (√(0.3)^{2} / √(0.2)^{2})

(0.3/0.2) ÷ (0.3/0.2) -1

(0.3/0.2 × 0.2/0.3) -1

1 – 1

0

**(v) **∛(0.1×0.1×0.1×13×13×13)

∛(0.1^{3}×13^{3})

0.1 × 13 = 1.3

**8. Show that:**

**(i) ∛ (729)/ ∛ (1000) = ∛ (729/1000)**

**(ii) ∛ (-512)/ ∛ (343) = ∛ (-512/343)**

**Solution:**

**(i) **∛ (729)/ ∛ (1000) = ∛ (729/1000)

Let us consider LHS ∛ (729)/ ∛ (1000)

∛ (729)/ ∛ (1000) = ∛ (9×9×9)/ ∛ (10×10×10)

** = **∛ (9^{3}/10^{3})

= 9/10

Let us consider RHS ∛ (729/1000)

∛ (729/1000) = ∛ (9×9×9/10×10×10)

** = **∛ (9^{3}/10^{3})

= 9/10

∴ LHS = RHS

**(ii) **∛ (-512)/ ∛ (343) = ∛ (-512/343)

Let us consider LHS ∛ (-512)/ ∛ (343)

∛ (-512)/ ∛ (343) = ∛-(8×8×8)/ ∛ (7×7×7)

** = **∛-(8^{3}/7^{3})

= -8/7

Let us consider RHS ∛ (-512/343)

∛ (-512/343) = = ∛-(8×8×8/7×7×7)

** = **∛-(8^{3}/7^{3})

= -8/7

∴ LHS = RHS

**9. Fill in the blanks:**

**(i) ∛(125×27) = 3 × …**

**(ii) ∛(8×…) = 8
(iii) ∛1728 = 4 × …**

**(iv) ∛480 = ∛3×2× ∛..**

**(v) ∛… = ∛7 × ∛8**

**(vi) ∛..= ∛4 × ∛5 × ∛6**

**(vii) ∛(27/125) = …/5**

**(viii) ∛(729/1331) = 9/…**

**(ix) ∛(512/…) = 8/13**

**Solution:**

**(i) **∛(125×27) = 3 × …

Let us consider LHS ∛(125×27)

∛(125×27) = ∛(5×5×5×3×3×3)

= ∛(5^{3}×3^{3})

= 5×3 or 3×5

**(ii) **∛(8×…) = 8

Let us consider LHS ∛(8×…)

∛(8×8×8) = ∛8^{3} = 8

**(iii) **∛1728 = 4 × …

Let us consider LHS

∛1728 = ∛(2×2×2×2×2×2×3×3×3)

= ∛(2^{3}×2^{3}×3^{3})

= 2×2×3

= 4×3

**(iv) **∛480 = ∛3×2× ∛..

Let us consider LHS

∛480 = ∛(2×2×2×2×2×3×5)

= ∛(2^{3}×2^{2}×3×5)

= ∛2^{3}× ∛3 × ∛2×2×5

= 2 × ∛3 × ∛20

**(v) **∛… = ∛7 × ∛8

Let us consider RHS

∛7 × ∛8 = ∛(7 × 8)

= ∛56

**(vi) **∛..= ∛4 × ∛5 × ∛6

Let us consider RHS

∛4 × ∛5 × ∛6 = ∛(4 × 5 × 6)

= ∛120

**(vii) **∛(27/125) = …/5

Let us consider LHS

∛(27/125) = ∛(3×3×3)/(5×5×5)

= ∛(3^{3})/(5^{3})

= 3/5

**(viii) **∛(729/1331) = 9/…

Let us consider LHS

∛(729/1331) = ∛(9×9×9)/(11×11×11)

= ∛(9^{3})/(11^{3})

= 9/11

**(ix) **∛(512/…) = 8/13

Let us consider LHS

∛(512/…) = ∛(2×2×2×2×2×2×2×2×2)

= ∛(2^{3}×2^{3}×2^{3})

= 2×2×2

= 8

So, 8/∛… = 8/13

when numerators are the same the denominators will also become equal.

8 × 13 = 8 × ∛…

∛…= 13

… = (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:**

The volume of a cubical box is 474.552 cubic metres

V = 8^{3},

Let ‘S’ be the side of the cube

8^{3} = 474.552 cubic metres

8 = ∛474.552

= ∛ (474552/1000)

Let us factorise 474552 into prime factors, we get:

474552 = 2×2×2×3×3×3×13×13×13

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

474552 = (2×2×2) × (3×3×3) × (13×13×13)

Now, ∛474.552 = ∛ ((2×2×2) × (3×3×3) × (13×13×13))

= 2×3×13

= 78

Also,

∛1000 = ∛ (10×10×10)

= ∛ (10)^{3}

= 10

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

8 = ∛ (474552/1000)

= 78/10

= 7.8

∴ 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 to be 2a, 3a, and 4a.

So according to the question:

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

8a^{3}+27 a^{3}+64 a^{3} = 0.334125

99a^{3} = 0.334125

a^{3} = 334125/1000000×99

= 3375/1000000

a = ∛ (3375/1000000)

= ∛((15×15×15)/ 100×100×100)

= 15/100

= 0.15

∴ The numbers are:

2×0.15 = 0.30

3×0.15 = 0.45

4×0.15 = 0.6

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

**Solution:**

Volume of the side s = 24389/216 = v

V= 8^{.3}

8 = ∛v

= ∛(24389/216)

By performing factorisation, we get,

= ∛(29×29×29/2×2×2×3×3×3)

= 29/(2×3)

= 29/6

∴ The length of the side is 29/6.

**13. Evaluate:**

**(i) ∛36 × ∛384**

**(ii) ∛96 × ∛144**

**(iii) ∛100 × ∛270**

**(iv) ∛121 × ∛297**

**Solution:**

**(i) **∛36 × ∛384

We know that ∛a × ∛b = ∛ (a×b)

By using the above formula, let us simplify

∛36 × ∛384 = ∛ (36×384)

The prime factors of 36 and 384 are

= ∛ (2×2×3×3) × (2×2×2×2×2×2×3×3×3)

= ∛(2^{3}×2^{3}×2^{3}×3^{3})

= 2×2×2×3

= 24

**(ii) **∛96 × ∛144

We know that ∛a × ∛b = ∛ (a×b)

By using the above formula, let us simplify

∛96 × ∛144 = ∛ (96×144)

The prime factors of 96 and 144 are

= ∛ (2×2×2×2×2×3) × (2×2×2×2×3×3)

= ∛(2^{3}×2^{3}×2^{3}×3^{3})

= 2×2×2×3

= 24

**(iii) **∛100 × ∛270

We know that ∛a × ∛b = ∛ (a×b)

By using the above formula, let us simplify

∛100 × ∛270 = ∛ (100×270)

The prime factors of 100 and 270 are

= ∛ (2×2×5×5) × (2×3×3×3×5)

= ∛(2^{3}×3^{3}×5^{3})

= 2×3×5

= 30

**(iv) **∛121 × ∛297

We know that ∛a × ∛b = ∛ (a×b)

By using the above formula, let us simplify

∛121 × ∛297 = ∛ (121×297)

The prime factors of 121 and 297 are

= ∛ (11×11) × (3×3×3×11)

= ∛(11^{3}×3^{3})

= 11×3

= 33

**14. Find the cube roots of the numbers 2460375, 20346417, 210644875, and 57066625 using the fact that
(i) 2460375 = 3375 × 729
(ii) 20346417 = 9261 × 2197
(iii) 210644875 = 42875 × 4913
(iv) 57066625 = 166375 × 343**

**Solution:**

**(i) **2460375 = 3375 × 729

By taking the cube root for the whole we get,

∛2460375 = ∛3375 × ∛729

Now perform factorisation

= ∛3×3×3×5×5×5 × ∛9×9×9

= ∛3^{3}×5^{3} × ∛9^{3}

= 3×5×9

= 135

**(ii)** 20346417 = 9261 × 2197

By taking the cube root for the whole we get,

∛20346417 = ∛9261 × ∛2197

Now, perform factorisation

= ∛3×3×3×7×7×7 × ∛13×13×13

= ∛3^{3}×7^{3} × ∛13^{3}

= 3×7×13

= 273

**(iii) **210644875 = 42875 × 4913

By taking the cube root for the whole we get,

∛210644875 = ∛42875 × ∛4913

Now perform factorisation

= ∛5×5×5×7×7×7 × ∛17×17×17

= ∛5^{3}×7^{3} × ∛17^{3}

= 5×7×17

= 595

**(iv) **57066625 = 166375 × 343

By taking the cube root for the whole we get,

∛57066625 = ∛166375 × ∛343

Now perform factorisation

= ∛5×5×5×11×11×11 × ∛7×7×7

= ∛5^{3}×11^{3} × ∛7^{3}

= 5×11×7

= 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}).

∴ 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}).

∴ 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, we are left with 571 only.

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

∴ 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 equal to 175(5^{3}<175<6^{3}).

∴ 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, by using the cube root table, we get,

∛7 = 1.913

∴ the answer is 1.913

**2. 70**

**Solution:**

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

We get,

∛70 = 4.121

∴ the answer is 4.121

**3. 700**

**Solution:**

700 = 70×10

By using the cube root table 700 will be in the column ∛10x against 70.

So we get,

∛700 = 8.879

∴ the answer is 8.879

**4. 7000**

**Solution:**

7000 = 70×100

∛7000 = ∛(7×1000) = ∛7 × ∛1000

By using the cube root table,

We get,

∛7 = 1.913

∛1000 = 10

∛7000 = ∛7 × ∛1000

= 1.913 × 10

= 19.13

∴ the answer is 19.13

**5. 1100**

**Solution:**

1100 = 11×100

∛1100 = ∛(11×100) = ∛11 × ∛100

By using the cube root table,

We get,

∛11 = 2.224

∛100 = 4.6642

∛1100 = ∛11 × ∛100

= 2.224 × 4.642

= 10.323

∴ the answer is 10.323

**6.780**

**Solution:**

780 = 78×10

By using the cube root table 780 would be in column ∛10x against 78.

We get,

∛780 = 9.205

**7. 7800**

**Solution:**

7800 = 78×100

∛7800 = ∛(78×100) = ∛78 × ∛100

By using the cube root table,

We get,

∛78 = 4.273

∛100 = 4.6642

∛7800 = ∛78 × ∛100

= 4.273 × 4.642

= 19.835

∴ the answer is 19.835

**8. 1346**

**Solution:**

Let us find the factors by using the factorisation method,

We get,

1346 = 2×673

∛1346 = ∛(2×676) = ∛2 × ∛673

Since, 670<673<680 = ∛670 < ∛673 < ∛680

By using the cube root table,

∛670 = 8.750

∛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) × 3 = 0.0132

∛673 = 8.750 + 0.013 = 8.763

∛1346 = ∛2 × ∛673

= 1.260 × 8.763

= 11.041

∴ the answer is 11.041

**9. 250**

**Solution:**

250 = 25×100

By using the cube root table 250 would be in column ∛10x against 25.

We get,

∛250 = 6.3

∴ the answer is 6.3

**10. 5112**

**Solution:**

Let us find the factors by using the factorisation method,

∛5112 = ∛2×2×2×3×3×71

= ∛2^{3}×3^{2}×71

= 2 × ∛3^{2} × ∛71

= 2 × ∛9 × ∛71

From the cube root table, we get,

∛9 = 2.080

∛71 = 4.141

∛5112 = 2 × ∛9 × ∛71

= 2 × 2.080 × 4.141

= 17.227

∴ the answer is 17.227

**11. 9800**

**Solution:**

∛9800 = ∛98 × ∛100

From the cube root table, we get,

∛98 = 4.610

∛100 = 4.642

∛9800 = ∛98 × ∛100

= 4.610 × 4.642

= 21.40

∴ the answer is 21.40

**12. 732**

**Solution:**

∛732

We know that the value of ∛732 will lie between ∛730 and ∛740

From the cube root table, we get,

∛730 = 9.004

∛740 = 9.045

By using the unitary method,

Difference between the values (740 – 730 = 10)

So, the difference in cube root values will be = 9.045 – 9.004 = 0.041

Difference between the values (732 – 730 = 2)

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

∛732 = 9.004+0.008 = 9.012

∴ the answer is 9.012

**13. 7342**

**Solution:**

∛7342

We know that the value of ∛7342 will lie between ∛7300 and ∛7400

From the cube root table, we get,

∛7300 = 19.39

∛7400 = 19.48

By using the unitary method,

Difference between the values (7400 – 7300 = 100)

So, the difference in cube root values will be = 19.48 – 19.39 = 0.09

Difference between the values (7342 – 7300 = 42)

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

∛7342 = 19.39+0.037 = 19.427

∴ the answer is 19.427

**14. 133100**

**Solution:**

∛133100 = ∛ (1331×100)

= ∛1331 × ∛100

= ∛ 11^{3} × ∛100

= 11 × ∛100

From the cube root table, we get,

∛100 = 4.462

∛133100 = 11 × ∛100

= 11 × 4.462

= 51.062

∴ the answer is 51.062

**15. 37800**

**Solution:**

∛37800

Firstly let us find the factors for 37800

∛37800 = ∛(2×2×2×3×3×3×175)

= ∛(2^{3}×3^{3}×175)

= 6 × ∛175

We know that the value of ∛175 will lie between ∛170 and ∛180

From the cube root table, we get,

∛170 = 5.540

∛180 = 5.646

By using the unitary method,

Difference between the values (180 – 170 = 10)

So, the difference in cube root values will be = 5.646 – 5.540 = 0.106

Difference between the values (175 – 170 = 5)

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

∛175 = 5.540 + 0.053 = 5.593

∛37800 = 6 × ∛175

= 6 × 5.593

= 33.558

∴ the answer is 33.558

**16. 0.27**

**Solution:**

∛0.27 = ∛(27/100) = ∛27/∛100

From the cube root table, we get,

∛27 = 3

∛100 = 4.642

∛0.27 = ∛27/∛100

= 3/4.642

= 0.646

∴ the answer is 0.646

**17. 8.6**

**Solution:**

∛8.6 = ∛(86/10) = ∛86/∛10

From the cube root table, we get,

∛86 = 4.414

∛10 = 2.154

∛8.6 = ∛86/∛10

= 4.414/2.154

= 2.049

∴ the answer is 2.049

**18. 0.86**

**Solution: **

∛0.86 = ∛(86/100) = ∛86/∛100

From the cube root table, we get,

∛86 = 4.414

∛100 = 4.642

∛8.6 = ∛86/∛100

= 4.414/4.642

= 0.9508

∴ the answer is 0.951

**19. 8.65**

**Solution:**

∛8.65 = ∛(865/100) = ∛865/∛100

We know that a value of ∛865 will lie between ∛860 and ∛870

From the cube root table, we get,

∛860 = 9.510

∛870 = 9.546

∛100 = 4.642

By using the unitary method,

Difference between the values (870 – 860 = 10)

So, the difference in cube root values will be = 9.546 – 9.510 = 0.036

Difference between the values (865 – 860 = 5)

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

∛865 = 9.510 + 0.018 = 9.528

∛8.65 = ∛865/∛100

= 9.528/4.642

= 2.0525

∴ the answer is 2.053

**20. 7532**

**Solution:**

∛7532

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

From cube root table we get,

∛7500 = 19.57

∛7600 = 19.66

By using the unitary method,

Difference between the values (7600 – 7500 = 100)

So, the difference in cube root values will be = 19.66 – 19.57 = 0.09

Difference between the values (7532 – 7500 = 32)

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

∛7532 = 19.57 + 0.029 = 19.599

∴ the answer is 19.599

**21. 833**

**Solution:**

∛833

We know that the value of ∛833 will lie between ∛830 and ∛840

From the cube root table, we get,

∛830 = 9.398

∛840 = 9.435

By using the unitary method,

Difference between the values (840 – 830 = 10)

So, the difference in cube root values will be = 9.435 – 9.398 = 0.037

Difference between the values (833 – 830 = 3)

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

∛833 = 9.398+0.011 = 9.409

∴ the answer is 9.409

**22. 34.2**

**Solution:**

∛34.2 = ∛(342/10) = ∛342/∛10

We know that the value of ∛342 will lie between ∛340 and ∛350

From the cube root table, we get,

∛340 = 6.980

∛350 = 7.047

∛10 = 2.154

By using the unitary method,

Difference between the values (350 – 340 = 10)

So, the difference in cube root values will be = 7.047 – 6.980 = 0.067

Difference between the values (342 – 340 = 2)

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

∛342 = 6.980 + 0.013 = 6.993

∛34.2 = ∛342/∛10

= 6.993/2.154

= 3.246

∴ 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 ‘a’cm

a^{3} = 275

a = ∛275

We know that the value of ∛275 will lie between ∛270 and ∛280

From the cube root table, we get,

∛270 = 6.463

∛280 = 6.542

By using the unitary method,

Difference between the values (280 – 270 = 10)

So, the difference in cube root values will be = 6.542 – 6.463 = 0.079

Difference between the values (275 – 270 = 5)

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

∛275 = 6.463 + 0.0395 = 6.5025

∴ the answer is 6.503cm

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