## MP Board Class 8th Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1

**MP Board Class 8 Maths Chapter 7 Exercise 7.1 Question 1.**

Which of the following numbers are not perfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Solution:

(i) 216 = __2 × 2 × 2__ × __3 × 3 × 3__

Since, all the prime factors of 216 appear in a group of three.

∴ 216 is a perfect cube.

(ii) 128 = __2 × 2 × 2__ × __2 × 2 × 2__ × 2

Since, all the prime factors of 128 doesn’t appear in a group of three.

∴ 128 is not a perfect cube.

(iii) 1000 = __2 × 2 × 2__ × __5 × 5 × 5__

Since, all the prime factors of 1000 appear in a group of three.

∴ 1000 is a perfect cube.

(iv) 100 = 2 × 2 × 5 × 5.

Since all the prime factors of 100 doesn’t appear in group of three.

∴ 100 is not a perfect cube.

(v) 46656 = __2 × 2 × 2__ × __2 × 2 × 2__ × __3 × 3 × 3__ × __3 × 3 × 3__

Since all the prime factors of 46656 appear in a group of three.

∴ 46656 is a perfect cube.

**Class 8 Maths Chapter 7 MP Board Question 2.**

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Solution:

(i) 243 = __3 × 3 × 3__ × 3 × 3

Clearly, the prime factor 3 doesn’t appear in a group of three.

∴ 243 is not a perfect cube.

So, to make 243 a perfect cube, we multiply it by 3.

In that case

243 × 3 = __3 × 3 × 3__ × __3 × 3 × 3__ = 729, which is a perfect cube.

(ii) 256 = __2 × 2 × 2__ × __2 × 2 × 2__ × 2 × 2

Clearly, the prime factor 2 doesn’t appear in a group of three.

256 is not a perfect cube.

So, to make 256 a perfect cube, we multiply it by 2.

In that case

256 × 2 = __2 × 2 × 2__ × __2 × 2 × 2__ × __2 × 2 × 2__ = 512, which is a perfect cube.

(iii) 72 = __2 × 2 × 2__ × 3 × 3

Clearly, the prime factor 3 doesn’t appear in a group of three.

∴ 72 is not a perfect cube.

So, to make 72 a perfect cube, we multiply it by 3.

In that case

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216, which is a perfect cube.

(iv) 675 = 3 × 3× 3 × 5 × 5

Clearly, the prime factor 5 doesn’t appear in a group of three.

∴ 675 is not a perfect cube.

So, to make 675 a perfect cube, we multiply it by 5.

In that case

675 × 5 = __3 × 3× 3__ × __5 × 5 × 5__

= 3375, which is a perfect cube.

(v) 100 = 2 × 2 × 5 × 5

Clearly, both the prime factors 2 and 5 doesn’t appear in a group of three.

∴ 100 is not a perfect cube.

So, to make 100 a perfect cube, we multiply it by 2 × 5 = 10.

In that case

100 × 2 × 5 = __2 × 2 × 2__ × __5 × 5 × 5__ = 1000, which is a perfect cube.

**MP Board Class 8 Maths Chapter 7 Question 3.**

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

(ii) 128

(iii) 135

(iv) 192

(v) 704.

Solution:

(i) 81 = __3 × 3 × 3__ × 3

Clearly, the prime factor 3 doesn’t appear in a group of three.

∴ 81 is not a perfect cube.

So, to make it a perfect cube, we must divide it by 3.

which is a perfect cube.

(ii) 128 = __2 × 2 × 2__ × __2 × 2 × 2__ × 2

Clearly, the prime factor 2 doesn’t appear in a group of three.

∴ 128 is not a perfect cube.

So, to make it a perfect cube, we must divide it by 2.

= 64,

which is a perfect cube.

(iii) 135 = __3 × 3 × 3__ × 5

Clearly, the prime factor 5 doesn’t appear in a group of three.

∴ 135 is not a perfect cube.

So, to make it a perfect cube, we must divide it by 5.

Thus the smallest number is 5 by which 135 must be divided.

(iv) 192 = __2 × 2 × 2__ × __2 × 2 × 2__ × 3

Clearly, the prime factor 3 doesn’t appear in a group of three.

∴ 192 is not a perfect cube.

So, to make it a perfect cube, we must divide it by 3.

which is a perfect cube.

(v) 704 = __2 × 2 × 2__ × __2 × 2 × 2__ × 11

Clearly, the prime factor 11 doesn’t appear in a group of three.

∴ 704 is not a perfect cube.

So, to make it a perfect cube, we must divide it by 11.

which is a perfect cube.

**Class 8 Maths Chapter 7 Exercise 7.1 Question 4.**

Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Solution:

We have cuboid of dimensions 2 × 5 × 5

Clearly, in above, the prime factors 2 and 5 both doesn’t appear in a group of three,

∴ To make it a perfect cube, we need to multiply it by 2 × 2 × 5.

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

= 1000, which is a perfect cube.

Thus, Prikshit need 20 cuboids to form a cube.