Question

Question 3(iv)
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube:
192

Answer 1

192
Prime factors of $192=2\times 2\times 2\times 2\times 2\times 2\times 3$
Here one factor 3 does not appear in a triplet.
Therefore, 192 must be divided by 3 to make it a perfect cube.
$\begin{array}{cc}2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$