(i) 243
Prime factors of 243 = 3 x 3 x 3 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
(ii) 256
Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 is required to make a 3’s group.
Therefore, 256 must be multiplied by 2 to make it a perfect cube.
(iii) 72
Prime factors of 72 = 2 x 2 x 2 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.
(iv) 675
Prime factors of 675 = 3 x 3 x 3 x 5 x 5
Here factor 5 does not appear in 3’s group.
Therefore 675 must be multiplied by 3 to make it a perfect cube.
(v) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here factor 2 and 5 both do not appear in 3’s group.
Therefore 100 must be multiplied by 2 x 5 = 10 to make it a perfect cube.

Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video

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https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/