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Class 8 RD Sharma Solutions – Chapter 4 Cubes and Cube Roots – Exercise 4.1 | Set 1
• Last Updated : 06 Apr, 2021

### Question 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

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

Cube of 40 is

40 = 40× 40× 40 = 64000

(vi) 55

Cube of 55 is

55 = 55× 55× 55 = 166375

(vii) 100

Cube of 100 is

100 = 100× 100× 100 = 1000000

(viii) 302

Cube of 302 is

302 = 302× 302× 302 = 27543608

(ix) 301

Cube of 301 is

301 = 301× 301× 301 = 27270901

### Question 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:

Finding the Cube of natural numbers up to 10

13 = 1 × 1 × 1 = 1

23 = 2 × 2 × 2 = 8

33 = 3 × 3 × 3 = 27

43 = 4 × 4 × 4 = 64

53 = 5 × 5 × 5 = 125

63 = 6 × 6 × 6 = 216

73 = 7 × 7 × 7 = 343

83 = 8 × 8 × 8 = 512

93 = 9 × 9 × 9 = 729

103 = 10 × 10 × 10 = 1000

Hence, we conclude

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

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

### Question 3. Observe the following pattern:

13 = 1

13 + 23 = (1+2)2

13 + 23 + 33 = (1+2+3)2

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

Solution:

From the given pattern,

13 + 23 + 33 +…+ 93

13 + 23 + 33 +…+ n3 = (1+2+3+…+n) 2

So when n = 10

13 + 23 + 33 +…+ 93 + 103 = (1+2+3+…+10) 2

= (55)2 = 55×55 = 3025

### Question 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:

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

So, let’s see how to find the cube of 3, 6, 9, 12 and 15

33 = 3 × 3 × 3 = 27

63 = 6 × 6 × 6 = 216

93 = 9 × 9 × 9 = 729

123 = 12 × 12 × 12 = 1728

153 = 15 × 15 × 15 = 3375

As we can see that all the cubes are divisible by 27

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

### Question 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 form i.e. when divided by 3 it leaves the remainder 1’

Solution:

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

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

43 = 4 × 4 × 4 = 64

73 = 7 × 7 × 7 = 343

103 = 10 × 10 × 10 = 1000

133 = 13 × 13 × 13 = 2197

163 = 16 × 16 × 16 = 4096

When all the above cubes are divided by ‘3’ leaves the remainder of 1.

Hence, the statement “The cube of a natural number of the form 3n+1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’ is true.

### Question 6. Write the cubes 5 natural numbers of 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 dividend by 3 the remainder is 2’

Solution:

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

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

53 = 5 × 5 × 5 = 125

83 = 8 × 8 × 8 = 512

113 = 11 × 11 × 11 = 1331

143 = 14 × 14 × 14 = 2744

173 = 17 × 17 × 17 = 4913

When all the above cubes are divided by ‘3’ leaves the remainder of 2

Hence, 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 dividend by 3 the remainder is 2’ is true.

### Question 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 73.

Solution:

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

So now, let us find the cube of 7, 14, 21, 28 and 35

73 = 7 × 7 × 7 = 343

143 = 14 × 14 × 14 = 2744

213 = 21× 21× 21 = 9261

283 = 28 × 28 × 28 = 21952

353 = 35 × 35 × 35 = 42875

We can see that all the above cubes are multiples of 73(343) as well.

Hence, the statement“The cube of a multiple of 7 is a multiple of 73 is true.

### Question 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

Finding the factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2 = 26 = (22)3 = 43

Hence, it’s a perfect cube.

(ii) 216

Finding the factors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33 = 63

Hence, it’s a perfect cube.

(iii) 243

Finding the factors of 243

243 = 3 × 3 × 3 × 3 × 3 = 35 = 33 × 32

Hence, it’s not a perfect cube.

(iv) 1000

Finding the factors of 1000

1000 = 2 × 2 × 2 × 5 × 5 × 5 = 23 × 53 = 103

Hence, it’s a perfect cube.

(v) 1728

Finding the factors of 1728

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 26 × 33 = (4 × 3 )3 = 123

Hence, it’s a perfect cube.

(vi) 3087

Finding the factors of 3087

3087 = 3 × 3 × 7 × 7 × 7 = 32 × 73

Hence, it’s not a perfect cube.

(vii) 4608

Finding the factors of 4608

4608 = 2 × 2 × 3 × 113

Hence, it’s not a perfect cube.

(viii) 106480

Finding the factors of 106480

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

Hence, it’s not a perfect cube.

(ix) 166375

Finding the factors of 166375

166375= 5 × 5 × 5 × 11 × 11 × 11 = 53 × 113 = 553

Hence, it’s a perfect cube.

(x) 456533

Finding the factors of 456533

456533= 11 × 11 × 11 × 7 × 7 × 7 = 113 × 73 = 773

Hence, it’s a perfect cube.

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

Solution:

(i) 216 = 23 × 33 = 63

It’s a cube of even natural number.

(ii) 512 = 29 = (23)3 = 83

It’s a cube of even natural number.

(iii) 729 = 33 × 33 = 93

It’s not a cube of even natural number.

(iv) 1000 = 103

It’s a cube of even natural number.

(v) 3375 = 33 × 53 = 153

It’s not a cube of even natural number.

(vi) 13824 = 29 × 33 = (23)3 × 33 = 83×33 = 243

It’s a cube of even natural number.

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

Solution:

(i) 125 = 5 × 5 × 5 × 5 = 53

It’s a cube of odd natural number.

(ii) 343 = 7 × 7 × 7 = 73

It’s a cube of odd natural number.

(iii) 1728 = 26 × 33 = 43 × 33 = 123

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

(iv) 4096 = 212 = (26)2 = 642

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

(v) 32768 = 215 = (25)3 = 323

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

(vi) 6859 = 19 × 19 × 19 = 193

It’s a cube of odd natural number.

### Question 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

Finding the factors of 675.

675 = 3 × 3 × 3 × 5 × 5

= 33 × 52

Hence, we need to multiply the product by 5.

(ii) 1323

Finding the factors of 1323

1323 = 3 × 3 × 3 × 7 × 7

= 33 × 72

Hence, we need to multiply the product by 7 to make it a perfect cube.

(iii) 2560

Finding the factors of 2560

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

= 23 × 23 × 23 × 5

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

(iv) 7803

Finding the factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 33 × 172

Hence, we need to multiply the product by 17 to make it a perfect cube.

(v) 107811

First find the factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 33 × 3 × 113

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

(vi) 35721

First find the factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 33 × 33 × 72

Hence, we need to multiply the product by 7 to make it a perfect cube.

### Chapter 4 Cubes and Cube Roots – Exercise 4.1 | Set 2

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