# RD Sharma Solutions Class 8 Cubes And Cube Roots Exercise 4.1

## RD Sharma Solutions Class 8 Chapter 4 Exercise 4.1

Q1. 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:

To find the cube of the given numbers

Cube of a number is defined as the number raised to the power 3.

(i) Cube of 7 = 73

= 7 x 7 x 7

Therefore, 73= 343

(ii) Cube of 12 = 123

= 12 x 12 x 12

Therefore, 123= 1728

(iii) Cube of 16 = 163

= 16 x 16 x 16

Therefore, 163 = 4096

(iv) Cube of 21 = 213

= 21 x 21 x 21

Therefore, 213= 9261

(v) Cube of 40 = 403

= 40 x 40 x 40

Therefore, 403= 64000

(vi) Cube of 55 = 553

= 55 x 55 x 55

Therefore, 553= 166375

(vii) Cube of 100 = 1003

= 100 x 100 x 100

Therefore, 1003 = 1000000

(viii) Cube of 302 = 3023

= 302 x 302 x 302

Therefore, 3023= 27543608

(ix) Cube of 301 = 3013

= 301 x 301 x 301

Therefore, 3013= 27270901

Q2. Write the cubes of all natural numbers between 1 and 0 and verify the following statements:

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

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

Solution:

The natural numbers between from 1 to 10 along with the cubes are listed and classified in the following table.

In order to check whether the cubes of the given number is even or odd, it is enough to check its divisibility by 2.

If the number is divided by 2, and the remainder will be zero, then it is an even number, otherwise, it will an odd number.

(i) From the given below table, it is evident that cubes of all the natural odd numbers are odd.

(ii) From the given below table, it is evident that cubes of all the natural even numbers are even.

 Number Cube Classification 1 1 Odd 2 8 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 3 27 Odd (Not an even number) 4 64 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 5 125 Odd (Not an even number) 6 216 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 7 343 Odd (Not an even number) 8 512 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 9 729 Odd (Not an even number) 10 1000 Even (Last digit is even, i.e., 0, 2, 4, 6, 8)

Q3. Observe the following pattern:

Write the next three rows and calculate the value of 13 + 23 + 33 + ……. + 93 + 103 by the above pattern.

Solution:

Expand the given pattern of numbers are as follows:

Therefore, the next three rows of the given number pattern is given below

Also, the required value of the given pattern is,

13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93 + 103 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) 2

= 552 = 3025

Hence the, the required value is 3025

Q4. 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 five natural numbers, that are multiples of 3 are: 3, 6, 9, 12 and 15.

So, the cubes of these five numbers are given as follows:

33 = 3 x 3 x 3 = 27

63 = 6 x 6 x 6 = 216

93 = 9 x 9 x 9 = 729

123 = 12 x 12 x 12 = 1728

153 = 15 x 15 x 15 = 3375

Now,write the cubes that are the multiples of 27. so, we have

27 = 27 x 1

216 = 27 x 8

729 = 27 x 27

1728 = 27 x 64

3375 = 27 x 125

It is observed that the cubes of the multiples of 3 also be written as the multiples of 27.

Hence, the cube of a natural number, which is a multiple of 3, is a multiple of 27 is verified.

Q5. 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 + I is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.

Solution:

Five natural numbers of the form (3n + 1) could be written by choosing n = 1, 2, 3, … etc.

Let five such numbers be 4, 7, 10, 13, and 16.

The cubes of these five numbers are: 43 = 64, 73 = 343, 103 = 1000, 133 =  2197 and l63 = 4096

The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as:

64 = 3 x 21 + 1,

It is of the form (3n + 1) for n = 21

343 = 3 x 114 + 1,

It is of the form (3n + 1) for n = 114

1000 = 3 x 333 + 1,

It is of the form (3n + 1) for n = 333

2197 = 3 x 732 + 1,

It is of the form (3n + 1) for n = 732

4096 = 3 x 1365 + 1,

It is of the form (3n + 1) for n = 1365

The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.

Q6. Write the cubes of 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 five natural numbers are of the form (3n + 2) and it can be written by choosing n = 1, 2, 3… etc.

So, the five such numbers are 5, 8, 11, 14, and 17.

When you the cubes of these five numbers, it becomes

53 = 125,

83 = 512,

113 = 1331,

143 = 2744,

173 = 4913.

Therefore, the cubes of the numbers 5, 8, 11, 14 and 17 are expressed as:

125 = 3 x 41 + 2,

It is of the form (3n + 2) for n = 41

512 = 3 x 170 + 2,

It is of the form (3n + 2) for n = 170

1331 = 3 x 443 + 2,

It is of the form (3n + 2) for n = 443

2744 = 3 x 914 + 2,

It is of the form (3n + 2) for n = 914

4913 = 3 x 1637 + 2,

It is of the form (3n + 2) for n = 1637

Therefore, form the above obsevations, we can say that the cubes of the numbers 5, 8, 11, 14, and 17 can also be expressed as the natural numbers of the form (3n + 2) for some natural number n.

Hence, the given statement is verified.

Q7. 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 73

Solution:

We know that the first five multiples of 7 can be written by choosing various values of a natural number “n”  using the expression 7n.

So. the five multiples be 7, 14, 21, 28 and 35.

Thus, the cubes of the above five numbers are:

73 = 343,

143 = 2744,

213 = 9261,

283 = 21952,

353 = 42875

Write the above obtained cubes numbers as a multiple of 73 as follows:

343 = 73 x l

2744 = 143 = 14 x 14 x 14

= (7 x 2) x (7 x 2) x (7 x 2)

= (7 x7 x 7) x (2 x 2 x 2)

= 73 x 23

9261 = 213 = 21 x 21 x 21

=(7 x 3) x (7 x 3) x (7 x 3)

= 73 x 33

21952 = 283 = 28 x 28 x 28

= (7 x 4) x (7 x 4) x (7 x 4)

= (7 x 7 x 7) x (4 x 4 x 4)

=73 x 43

42875 = 353 = 35 x 35 x 35

= (7 x 5) x (7 x 5) x (7 x 5)

= (7 x 7 x 7) x (5 x 5 x 5)

= 73 x 53

It is observed that the cube of multiple of 7 is a multiple of 73.

Hence, the given statement is verified.

Q8. 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) Factorise the number 64 into prime factors,

We get,

64 = 2 x 2 x 2 x 2 x 2 x 2

Now, group the factors in triples of equal factors:

64 = {2 x 2 x 2} x {2 x 2 x 2}

So, it is observed that the prime factors of 64 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 64 is a perfect cube.

(ii) Factorise the number 216 into prime factors,

We get,

216 = 2 x 2 x 2 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

216 = {2 x 2 x 2} x {3 x 3 x 3}

So, it is observed that the prime factors of 216 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 216 is a perfect cube.

(iii) Factorise the number 243 into prime factors,

We get,

243 = 3 x 3 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

243 = {3 x 3 x 3} x 3 x3

So, it is observed that the prime factors of 243 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 243 is not a perfect cube.

(iv) Factorise the number 1000 into prime factors,

We get,

1000 = 2 x 2 x 2 x 5 x 5 x 5

Now, group the factors in triples of equal factors:

1000 = {2 x 2 x 2} x {5 x 5 x 5}

So, it is observed that the prime factors of 1000 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 1000 is a perfect cube

(v) Factorise the number 1728 into prime factors,

We get,

1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

1728 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3}

So, it is observed that the prime factors of 1728 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 1728 is a perfect cube

(vi) Factorise the number 3087 into prime factors,

We get,

3087 = 3 x 3 x 7 x 7 x 7

Now, group the factors in triples of equal factors:

3087 = 3 x 3 x {7 x 7 x 7}

So, it is observed that the prime factors of 3087 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 3087 is not a perfect cube

(vii) Factorise the number4608 into prime factors,

We get,

4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Now, group the factors in triples of equal factors:

4608 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x 3 x 3

So, it is observed that the prime factors of 4608 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 4608 is not a perfect cube

(viii) Factorise the number106480 into prime factors,

We get,

106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11

Now, group the factors in triples of equal factors:

106480 = {2 x 2 x 2} x 2 x 5 x {11 x 11 x 11}

So, it is observed that the prime factors of 106480 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 106480 is not a perfect cube

(ix) Factorise the number166375 into prime factors,

We get,

166375 = 5 x 5 x 5 x 11 x 11 x 11

Now, group the factors in triples of equal factors:

166375 = {5 x 5 x 5} x {11 x 11 x 11}

So, it is observed that the prime factors of 166375 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 166375 is a perfect cube

(x) Factorise the number 456533 into prime factors,

We get,

456533 = 7 x 7 x 7 x 11 x 11 x11

Now, group the factors in triples of equal factors:

456533={7 x 7 x 7} x {11 x 11 x 11}

So, it is observed that the prime factors of 456533 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 456533 is a perfect cube

Q9. Which of the following are cubes of even natural numbers?

216, 512, 729, 1000, 3375, 13824

Solution:

Given:

We know that,

the cubes of all even natural numbers are even.

The numbers which are the cubes of even natural numbers are: 216, 512, 1000 and 13824

These numbers are even and it can ve verified using the divisibility rule test of 2.

It means that if a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.

Hence, the cubes of even natural numbers are 216, 512, 1000 and 13824.

Q10. Which of the following are cubes of odd natural numbers?

125, 343, 1728, 4096, 32768, 6859

Solution:

Given:

We know that,

the cubes of all odd natural numbers are odd.

The numbers which are the cubes of odd natural numbers are: 125, 343, and 6859

These numbers are odd and it can ve verified using the divisibility rule test of 2.

It means that if a number is divisible by 2, it is even number. i.e., if the number ends with 0, 2, 4, 6 or 8. Otherwise, it is an odd number.

Hence, the cubes of odd natural numbers are 125, 343, and 6859

Q11. 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) Factorise the number 675 into prime factors,

We get,

675 = 3 x 3 x 3 x 5 x 5

Now, group the factors in triples of equal factors:

675 = {3 x 3 x 3} x 5 x 5

So, it is observed that the prime factors of 675 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 675 is not a perfect cube

However, if the number 675 is multiplied by 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 5 should be multiplied with 675 to make it a perfect cube.

(ii) Factorise the number 1323 into prime factors,

We get,

1323 =  3 x 3 x 3 x 7 x 7

Now, group the factors in triples of equal factors:

1323 = {3 x 3 x 3} x 7 x 7

So, it is observed that the prime factors of 1323 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence,1323 is not a perfect cube

However, if the number 1323 is multiplied by 7, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 7 should be multiplied with 1323 to make it a perfect cube.

(iii)  Factorise the number 2560 into prime factors,

We get,

2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5

Now, group the factors in triples of equal factors:

2560 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x 5

So, it is observed that the prime factors of 2560 can be grouped into triples of equal factors

And, also, there is 1 factor left over.

Hence, 2560  is not a perfect cube

However, if the number 2560 is multiplied by 5 x 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 25 should be multiplied with 2560 to make it a perfect cube.

(iv) Factorise the number 7803 into prime factors,

We get,

7803 = 3 x 3 x 3 x 17 x 17

Now, group the factors in triples of equal factors:

7803 = {3 x 3 x 3} x 17 x 17

So, it is observed that the prime factors of 7803 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 7803  is not a perfect cube

However, if the number 7803 is multiplied by 17, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 17 should be multiplied with 7803 to make it a perfect cube.

(v) Factorise the number 107811  into prime factors,

We get,

107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Now, group the factors in triples of equal factors:

107811 ={3 x 3 x 3} x 3 x{11 x 11 x 11}

So, it is observed that the prime factors of 107811 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 107811  is not a perfect cube

However, if the number 107811 is multiplied by 3 x 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 9 should be multiplied with 107811 to make it a perfect cube.

(vi)  Factorise the number 35721 into prime factors,

We get,

35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Now, group the factors in triples of equal factors:

35721 = {3 x 3 x 3} x {3 x 3 x 3} x 7 x 7

So, it is observed that the prime factors of 35721 can be grouped into triples of equal factors

And, also, there is 1 factor left over.

Hence, 35721  is not a perfect cube

However, if the number 35721  is multiplied by 7, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 7 should be multiplied with 35721 to make it a perfect cube.

Q12. 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) Factorise the number 675 into prime factors,

We get,

675 = 3 x 3 x 3 x 5 x 5

Now, group the factors in triples of equal factors:

675 = {3 x 3 x 3} x {5 x 5}

So, it is observed that the prime factors of 675 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 675  is not a perfect cube

However, if the number 675  is divided by 5 x 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 675 should be divided by 25 to make it a perfect cube.

(ii) Factorise the number 8640 into prime factors,

We get,

8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5

Now, group the factors in triples of equal factors:

8640 = {2 x 2 x 2} x {2 x 2 x 2 } x {3 x 3 x 3} x 5

So, it is observed that the prime factors of 8640 can be grouped into triples of equal factors

And, also, there is 1 factor left over.

Hence, 8640  is not a perfect cube

However, if the number 8640  is divided by 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 8640 should be divided by 5 to make it a perfect cube.

(iii) Factorise the number 1600 into prime factors,

We get,

1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

Now, group the factors in triples of equal factors:

1600 = {2 x 2 x 2} x {2 x 2 x 2} x 5 x 5

So, it is observed that the prime factors of 1600 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 1600  is not a perfect cube

However, if the number 1600  is divided by 5 x 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 1600 should be divided by 25 to make it a perfect cube.

(iv) Factorise the number 8788 into prime factors,

We get,

8788 = 2 x 2 x 13 x 13 x 13

Now, group the factors in triples of equal factors:

8788 = 2 x 2 x {13 x 13 x 13}

So, it is observed that the prime factors of 8788 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 8788  is not a perfect cube

However, if the number 8788  is divided by 2 x 2, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 8788 should be divided by 4 to make it a perfect cube.

(v) Factorise the number 7803 into prime factors,

We get,

7803 = 3 x 3 x 3 x 17 x 17

Now, group the factors in triples of equal factors:

7803 = {3 x 3 x 3} x 17 x 17

So, it is observed that the prime factors of 7803 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 7803  is not a perfect cube

However, if the number 7803  is divided by 17 x 17, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 7803 should be divided by 289 to make it a perfect cube.

(vi) Factorise the number 107811 into prime factors,

We get,

107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Now, group the factors in triples of equal factors:

{3 x 3 x 3} x 3 x {11 x 11 x 11}

So, it is observed that the prime factors of 107811 can be grouped into triples of equal factors

And, also, there is 1 factor left over.

Hence, 107811  is not a perfect cube

However, if the number 107811  is divided by 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 107811 should be divided by 3 to make it a perfect cube.

(vii) Factorise the number 35721 into prime factors,

We get,

35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Now, group the factors in triples of equal factors:

35721 = {3 x 3 x 3} x {3 x 3 x 3} x 7 x7

So, it is observed that the prime factors of 35721 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 35721  is not a perfect cube

However, if the number 35721  is divided by 7 x7, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 35721  should be divided by 49 to make it a perfect cube.

(viii) Factorise the number 243000 into prime factors,

We get,

243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

Now, group the factors in triples of equal factors:

243000 = {2 x 2 x 2} x {3 x 3 x 3} x 3 x 3 x {5 x 5 x 5}

So, it is observed that the prime factors of 243000 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 243000  is not a perfect cube

However, if the number 243000  is divided by 3 x 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 243000 should be divided by 9 to make it a perfect cube.

Q13. Prove that if a number is trebled they its cube is 27 times the cube of the given number.

Solution:

Let us assume a number be “n”.

So, the cube of a number is n3.

When the number n is trebled,

i.e., 3n, its cube becomes

(3n)3 = 3n x n3

= 27 n3

It is observed that the cube of 3n is 27 times the cube of the number “n”.

Hence, the given statement is proved.

Q14. What happens to the cube of a number if the number is multiplied by

(i) 3? (ii) 4? (iii) 5?

Solution:

(i) Let us assume a number be “n”.

So, the cube of a number is n3.

When the number n is multiplied by 3,

i.e., 3n, its cube becomes

(3n)3 = 3n x n3

= 27 n3

Therefore, it is observed that the cube of 3n is 27 times the cube of the number “n”.

Hence, if a number is multiplied by 3,  then the cube is 27 times the cube of that number.

(ii)  Let us assume a number be “n”.

So, the cube of a number is n3.

When the number n is multiplied by 4,

i.e., 4n, its cube becomes

(4n)3 = 3n x n3

= 64 n3

Therefore, it is observed that the cube of 4n is 46 times the cube of the number “n”.

Hence, if a number is multiplied by 4,  then the cube is 64 times the cube of that number.

(iii)  Let us assume a number be “n”.

So, the cube of a number is n3.

When the number n is multiplied by 5,

i.e., 5n, its cube becomes

(5n)3 = 3n x n3

= 125 n3

Therefore, it is observed that the cube of 5n is 125 times the cube of the number “n”.

Hence, if a number is multiplied by 5,  then the cube is 125 times the cube of that number.

Q15. Find the volume of a cube, one face of which has an area of 64 m2.

Solution:

Given that. the area of a cube face is 64 m2

We know that A = a2 Square units

where “a” = Side of the cube

Also, the volume of a cube is given by:

V = a cubic units

To find the side of a cube, substitute the given values in the area formula,

s2 = 64

=>  s = √64= 8 m

Therefore, the side of the cube = 8m

Now, the volume  of a cube is given by:

V = s3 = 83

V = 8 x 8 x 8 = 512 m3

Hence, the volume of the cube is 512 m3.

Q16. Find the volume of a cube whose surface area is 384 m2.

Solution:

Given that,

The surface area of a cube, SA = 384 m3.

where a = Side of the cube

Also, the volume of a cube is given by:

V = a cubic units

We know that the surface area of a cube, SA = 6asquare units

Substitute the given values in the above formula, we get,

6a=384

a2= 384 /6

Therefore, a = 8 m

Now, substitute the value of “a” in volume formula, we get,

V = 8 cubic units

V = 8 x 8 x 8

V =512

Therefore, the volume of a cone = 512 m2

Q17. Evaluate the following:

(i) {(52 + 122)1/2}3 (ii) {(62 + 82)1/2}3

Solution:

(i) Given expression: {(52 + 122)1/2}3

T0 evaluate the expression, proceed with the steps,

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

= {√169}3

= (√(13 x13))3

Cancel square and square root, it becomes (13)3

= 13 x13 x13 = 2197

Therefore, {(52 + 122)1/2}3= 2197

(ii) Given expression:{(62 + 82)1/2}3

T0 evaluate the expression, proceed with the steps,

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

= {√100}3

= (√(10 x10))3

Cancel square and square root, it becomes (10)3

= 10 x10 x10 = 1000

Therefore, {(62 + 82)1/2}3 = 1000

Q18. Write the units digit of the cube of each of the following numbers: 31, 109, 388, 833, 4276, 5922, 77774, 44447, 125125125

Solution:

We know that the propertie of cube numbers are:

When numbers end with digits 1, 4, 5, 6 or 9, then its cube will end with the same digit.

When a number ends with 2, then its cube will end with 8.

When a number ends with 8, then its cube will end with 2.

When a number ends with 3, then its cube will end with 7.

When a number ends with 7, then its cube will end with 3.

From the above-given proper properties, we can say that,

31 = Cube of the number 31 ends with 1.

109 = Cube of the number 109 ends with 9.

388 = Cube of the number 388 ends with 2.

833 = Cube of the number 833 ends with 7.

4276 = Cube of the number 4276 ends with 6.

5922  = Cube of the number 5922 ends with 8.

77774 = Cube of the number 77774 ends with 4.

44447 = Cube of the number 44447 ends with 3.

125125125 = Cube of the number 125125125 ends with 5.

Q19. Find the cubes of the following numbers by column method:

(i) 35 (ii) 56 (iii) 72

Solution:

(i) To find the cube of a number 35 using the column method, proceed with the steps

Let a = 3 and b = 5

 Column I a3 Column II 3 x a2 x b Column III 3 x a x b2 Column IV b3 33 = 27 3 x a2 x b = 3 x 32 x 5 = 135 3 x a x b2 = 3 x 3 x 52 = 225 53 = 125 + 15 + 23 + 12 125 42 158 237 42 8 7 5

Therefore, the cube of 35 is 42875.

(ii) To find the cube of a number 56 using the column method, proceed with the steps

Let a = 5 and b = 6

 Column I a3 Column II 3 x a2 x b Column III 3 x a x b2 Column IV b3 53 = 125 3 x a2 x b = 3 x 52 x 6 = 450 3 x a x b2 = 3 x 5 x 62 = 540 63 = 216 + 50 + 56 + 21 216 175 506 561 175 6 1 6

Therefore, the cube of 56 is 175616

(iii) To find the cube of a number 72 using the column method, proceed with the steps

Let a = 7 and b = 2

 Column I a3 Column II 3 x a2 x b Column III 3 x a x b2 Column IV b3 73 3 x a2 x b = 3 x 72 x 2 = 294 3 x a x b2 = 3 x 7 x 22 = 84 23 = 8 + 30 + 8 + 0 8 373 302 84 373 2 4 8

Therefore, the cube of 72 is 373248

Q20. Which of the following numbers are not perfect cubes?

(i) 64 (ii) 216 (iii) 243 (iv) 1728

Solution:

(i)  Factorise the number 64 into prime factors,

We get,

64 = 2 x 2 x 2 x 2 x 2 x 2

Now, group the factors in triples of equal factors:

{2 x 2 x 2} x {2 x 2 x 2}

So, it is observed that the prime factors of 64 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 64  is a perfect cube

(ii) Factorise the number 216 into prime factors,

We get,

216 = 2 x 2 x 2 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

216 = {2 x 2 x 2} x {3 x 3 x 3}

So, it is observed that the prime factors of 216 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 216 is a perfect cube

(iii)  Factorise the number 243 into prime factors,

We get,

243 = 3 x 3 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

243 = {3 x 3 x 3} x 3 x 3

So, it is observed that the prime factors of 243 can be grouped into triples of equal factors

And, also there are 2 factors left over.

Hence, 243 is not a perfect cube

(iv) Factorise the number 1728 into prime factors,

We get,

1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

1728 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3}

So, it is observed that the prime factors of 1728 can be grouped into triples of equal factors

And, also, there is no factor left over.

Hence, 1728 is a perfect cube

Therefore, (iii) 243 is the required number, which is not a perfect cube.

Q21. 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 non-perfect cube in Q, No. 20 is 243.

Therefore, the smallest number which is obtained by multiplying and diving to get a perfect cube as follows:

(a) Factorise the number 243 into prime factors,

We get,

243 = 3 x 3 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

243 ={3 x 3 x 3}x 3 x 3

So, it is observed that the prime factors of 243 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 243  is not a perfect cube

However, if the number 243 is multiplied by 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 3 should be multiplied with 243 to make it a perfect cube.

(b) Factorise the number 243 into prime factors,

We get,

243 = 3 x 3 x 3 x 3 x 3

Now, group the factors in triples of equal factors:

243 = {3 x 3 x 3} x 3 x 3

So, it is observed that the prime factors of 243 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 243  is not a perfect cube

However, if the number 243  is divided by 3 x 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.

Therefore, the number 243 should be divided by 9 to make it a perfect cube.

Q.22: By taking three different values of n, verify the truth of the following statements:

(i) If n is even, then n3 is also even.

(ii) if n is odd, then n3 is also odd.

(iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as the remainder when divided by 3.

(iv) If a natural number n is of the form 3p + 2 then n3 also a number of the same type.

Solution:

(i) consider three even natural numbers: 2, 4 and 8.

The cubes of the numbers are:

23 = 8,

43 = 64,

83 = 512

With the help of the divisibility test, it is observed that 8, 64 and 512 are divisible by 2.

Therefore, the numbers given are even.

Hence, it verifies the statement.

(ii) consider the three odd natural numbers: 3, 9 and 27.

The cubes of the above-given numbers are:

33 = 27,

93 = 729,

273 = 19683

With the help of the divisibility test, it is observed that 27, 729 and 19683 are divisible by 3.

Thus, the numbers are odd.

Hence, it verifies the statement.

(iii) Given that, the three natural numbers of the form (3n + 1) is written by choosing n = 1,2,3… etc.

Now, consider the three numbers such as 4,7 and 10.

The cubes of the three chosen numbers are:

43 = 64,

73 = 343

103 = 1000

So, the cubes of 4,7 and 10 can written as:

64 = 3 x 21 + 1,

It is of the form (3n + 1) for n = 21

343 = 3 x 114 + 1,

It is of the form (3n + 1) for n = 114

1000 = 3 x 333 + 1,

It is of the form (3n + 1) for n = 333

Therefore, the cubes of 4, 7, and 10 can be written as the natural numbers of the form (3n + 1) for some natural number n.

Hence, the given statement is verified.

(iv)  Given that, the three natural numbers of the form (3p + 2) can be expressed by choosing p = 1,2,3… etc.

Now, consider the three numbers such as 5, 8 and 11.

Therefore, the cubes of the three chosen numbers are:

53 = 125,

83 = 512

113 = 1331

So, the cubes of 5, 8, and 11 can be expressed as:

125 = 3 x 41 + 2,

It is of the form (3p + 2) for p = 41

512 = 3 x 170 + 2,

It is of the form (3p + 2) for p = 170

1331 = 3 x 443 + 2,

It is of the form (3p + 2) for p = 443

Therefore, the cubes of 5, 8, and 11 could be written as the natural numbers of the form (3p + 2) for some natural number p.

Hence, the given statement is verified.

Q23. 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, a3 is always greater than a2.

(vi) If a and b are integers such that a2 > b2, then a3 > b3.

(vii) if a divides b, then a3 divides b3

(viii) If a2 ends in 9, then a3 ends in 7.

(ix) If a2 ends in 5, then a3 ends in 25.

(x) I a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.

Solution:

(i) False

Factorise the number 392 into prime factors,

We get,

392 = 2 x 2 x 2 x 7 x 7

Now, group the factors in triples of equal factors:

392 = {2 x 2 x 2} x 7 x 7

So, it is observed that the prime factors of 392 can be grouped into triples of equal factors

And, also, there are 2 factors left over.

Hence, 392 is not a perfect cube

(ii) True

Factorise the number 8640 into prime factors,

We get,

8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5

Now, group the factors in triples of equal factors:

8640 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3} x 5

So, it is observed that the prime factors of 8640 can be grouped into triples of equal factors

And, also, there is 1 factor left over.

Hence, 8640 is not a perfect cube

(iii) True

It is noted that perfect cube always ends with multiples of 3 zeros,

It means that, it will always ends with 3 zeros, 6 zeros etc.

(iv) False.

Since the unit digit of 64 is 4, the number 64 is a perfect cube

(v) False

The given statement is not true for a negative integer

For Example: (-3)2 = 9; (-3)3 = – 27

=> (-3)3 < (-3)2

(vi) False

The given statement is not true for negative integers.

For Example: (-6)2 > (-5)2 but (-6)3 < (-5)3

(vii) True

For the given condition, a divides b

$\frac{b^{3}}{a^{3}} = \frac{b \times b \times b}{a \times a \times a} = \frac{(ax) \times (ax) \times (ax)}{a \times a \times a}$

a divides b

b = ax for some x

$\frac{b^{3}}{a^{3}} = \frac{(ax) \times (ax) \times (ax)}{a \times a \times a} = x^{3} \\ => b^{3} = a^{3}(x^{3}) a^{3} \; divides \; b^{3}$

(viii) False

It is known that, a3 ends in 7 if “a” ends with 3.

So for every a2 ending in 9, it is not necessary that a is 3.

For example, if 49 is a square of 7, then the cube of 7 is 343.

(ix) False

Example : 352 = 1225 but 353 = 42875

(x) False

For example: 1002 = 10000 and 1003 = 100000