NCERT Class 8 Mathematics Solutions: Chapter 7 – Cubes and Cube Roots Exercise 7.1 Part 2 (For CBSE, ICSE, IAS, NET, NRA 2023)

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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
3243
381
327
39
33
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
2256
2128
264
232
216
28
24
22
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
272
236
218
39
33
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
3675
3225
375
525
55
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
2100
250
525
55
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.