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) 192 (v) 704

(i) 81
Prime factors of 81 = 3 x 3 x 3 x 3
Here one factor 3 is not grouped in triplets.
Therefore 81 must be divided by 3 to make it a perfect cube.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 must be divided by 2 to make it a perfect cube.
(iii) 135
Prime factors of 135 = 3 x 3 x 3 x 5
Here one factor 5 does not appear in a triplet.
Therefore, 135 must be divided by 5 to make it a perfect cube.
(iv) 192
Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Here one factor 3 does not appear in a triplet.
Therefore, 192 must be divided by 3 to make it a perfect cube.
(v) 704
Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.

Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video

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