Question

3. Find the smallest number by which each of the following numbers must be dividing
obtain a perfect cube.
(i) 81
(ii) 128
(i) 135
(iv) 192
(v) 704​

Answer 1

Step-by-step explanation:

(i)81

Prime factors of 81 = 3\times3\times3\times33×3×3×3

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\times2\times2\times2\times2\times22×2×2×2×2×2

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\times3\times3\times53×3×3×5

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\times2\times2\times2\times2\times32×2×2×2×2×3

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\times2\times2\times2\times2\times2\times112×2×2×2×2×2×11

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a perfect cube.

Answered by B.Muthu Manikandan