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.