NCERT Class 8 Mathematics Solutions: Chapter 7 – Cubes and Cube Roots Exercise 7.1 Part 2

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Image of cube roots 9261 by prime factorisation method

Image of Cube Roots 9261 by Prime Factorisation Method

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

Answer:

(i) 243

Table shows cube roots of 243 by prime factorization method

3

243

3

81

3

27

3

9

3

3

1

Prime factors of

To make 243 a cube, one more 3 is required.

Hence, the number 243 multiply by 3 to obtain a perfect cube.

(ii) 256

Table shows cube roots of 256 by prime factorization method

2

256

2

128

2

64

2

32

2

16

2

8

2

4

2

2

1

Prime factors of 256 =

To make 256 a cube, one more 2 is required.

Hence, the number 256 multiply by 2 to obtain a perfect cube.

(iii) 72

Table shows cube roots of 72 by prime factorization method

2

72

2

36

2

18

3

9

3

3

1

Prime factors of

To make 72 a cube, one more 3 is required.

Hence, the number 72 multiply by 3 to obtain a perfect cube.

(iv) 675

Table shows cube roots of 675 by prime factorization method

3

675

3

225

3

75

5

25

5

5

1

Prime factors of

To make 675 a cube, one more 5 is required.

Hence, the number 675 multiply by 5 to obtain a perfect cube.

(v) 100

Table shows cube roots of 100 by prime factorization method

2

100

2

50

5

25

5

5

1

Prime factors of

To make 100 a cube, we require one more 2 and one more 5.

Hence, the number 100 multiply by to obtain a perfect cube.