# Mathematics – Class 8 – Chapter 7 – Cubes and cube root – Exercise 7.1 – NCERT Exercise Solution

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.

Hence, 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.

Hence, 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.

Hence, 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.

Hence, 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.

Hence, 46656 is a perfect cube.

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

(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

Hence, the required smallest number to be multiplied = 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

Hence, the required smallest number to be multiplied = 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

Hence, the required smallest number to be multiplied = 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

Hence, the required smallest number= 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

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

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:

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,

81 ÷ 3 = 27 which is a perfect cube.

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

(ii)

Prime factorisation of 128:

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,

128 ÷ 2 = 64 which is a perfect cube.

Hence, 2 is the required smallest number.

(iii)

Prime factorisation of 135:

135 = 3 × 3 × 3 × 5 = 33 × 5

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

135 ÷ 5 = 27 which is a perfect cube.

Hence, 5 is the required smallest number.

(iv)

Prime factorisation of 192:

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,

192 ÷ 3 = 64 which is a perfect cube.

Hence, 3 is the required smallest number.

(v)

Prime factorisation of 704:

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,

704 ÷ 11 = 64 which is a perfect cube.

Hence, 11 is the required smallest number.

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:

Given, the sides of the cuboid as 5 cm, 2 cm and 5 cm.

Now,

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

By the prime factorisation of 50, we get

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

Hence, the required number of cuboids = 20.

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