## NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

**Ex 7.1 Class 8 Maths Question 1.**

**Which of the following numbers are not perfect cubes?**

**(i) 216**

**(ii) 128**

**(iii) 1000**

**(iv) 100**

**(v) 46656**

**Solution:**

(i) Prime factorisation of 216 is:

216 = 2 × 2 × 2 × 3 × 3 × 3

In the above factorisation, 2 and 3 have formed a group of three.

Thus, 216 is a perfect cube.

(ii) Prime factorisation of 128 is:

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, 2 is left without making a group of three.

Thus 128 is not a perfect cube.

(iii) Prime factorisation of 1000, is:

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, no number is left for making a group of three.

Thus, 1000 is a perfect cube.

(iv) Prime factorisation of 100, is:

100 = 2 × 2 × 5 × 5

Here 2 and 5 have not formed a group of three.

Thus, 100 is not a perfect cube.

(v) Prime factorisation of 46656 is:

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here 2 and 3 have formed the groups of three.

Thus, 46656 is a perfect cube.

**Ex 7.1 Class 8 Maths 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) Prime factorisation of 243, is:

243 = 3 × 3 × 3 × 3 × 3 = 3³ × 3 × 3

Here, number 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3

Thus, the required smallest number to be multiplied is 3.

(ii) Prime factorisation of 256, is:

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2 × 2

Here, a number 2 is needed to make 2 × 2 a group of three, i.e., 2 × 2 × 2

Thus, the required smallest number to be multiplied is 2.

(iii) Prime factorisation of 72, is:

72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3 × 3

Here, a number 3 is required to make 3 × 3 a group of three, i.e. 3 × 3 × 3

Thus, the required smallest number to be multiplied is 3.

(iv) Prime factorisation of 675, is:

675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5 × 5

Here, a number 5 is required to make 5 × 5 a group of three to make it a perfect cube, i.e. 5 × 5 × 5

Thus, the required smallest number is 5.

(v) Prime factorisation of 100, is:

100 = 2 × 2 × 5 × 5

Here, number 2 and 5 are needed to multiplied 2 × 2 × 5 × 5 to make it a perfect cube, i.e., 2 × 2 × 2 × 5 × 5 × 5

Thus, the required smallest number to be multiplied is 2 × 5 = 10.

**Ex 7.1 Class 8 Maths 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) 92**

**(v) 704**

**Solution:**

(i) Prime factorisation of 81, is:

81 = 3 × 3 × 3 × 3 = 3³ × 3

Here, a number 3 is the number by which 81 is divided to make it a perfect cube,

i.e., 81 ÷ 3 = 27 which is a perfect cube.

Thus, the required smallest number to be divided is 3.

(ii) Prime factorisation of 128, is:

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2

Here, a number 2 is the smallest number by which 128 is divided to make it a perfect cube,

i.e., 128 ÷ 2 = 64 which is a perfect cube.

Thus, 2 is the required smallest number.

(iii) Prime factorisation of 135 is:

135 = 3 × 3 × 3 × 5 = 3³ × 5

Here, 5 is the smallest number by which 135 is divided to make a perfect cube,

i.e., 135 ÷ 5 = 27 which is a perfect cube.

Thus, 5 is the required smallest number.

(iv) Prime factorisation of 192 is:

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2³ × 2³ × 3

Here, 3 is the smallest number by which 192 is divided to make it a perfect cube,

i.e., 192 ÷ 3 = 64 which is a perfect cube.

Thus, 3 is the required smallest number.

(v) Prime factorisation of 704 is:

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 2³ × 2³ × 11

Here, 11 is the smallest number by which 704 is divided to make it a perfect cube,

i.e., 704 ÷ 11 = 64 which is a perfect cube.

Thus, 11 is the required smallest number.

**Ex 7.1 Class 8 Maths Question 4.**

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

**Solution:**

The sides of the cuboid are given as 5 cm, 2 cm and 5 cm.

Volume of the cuboid = 5 cm × 2 cm × 5 cm = 50 cm³

For the prime factorisation of 50, we have

50 = 2 × 5 × 5

To make it a perfect cube, we must have

2 × 2 × 2 × 5 × 5 × 5

= 20 × (2 × 5 × 5)

= 20 × volume of the given cuboid

Thus, the required number of cuboids = 20.