Question

Find the smallest number(s) by which $3087,33275,120393$ must be divided respectively so that quotient is a perfect cube.

Answer 1

Consider $3087$

$\begin{array}{cc}3& 3087\\ 3& 1029\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$\Rightarrow 3087=3^3\times 7^3$

Hence, $3087$ must be divided by $9$ to make it a perfect cube.

$\therefore \frac{3087}{9}=343=7^3$

Consider $33275$

$\begin{array}{cc}5& 33275\\ 5& 6655\\ 11& 1331\\ 11& 121\\ 11& 11\\ & 1\end{array}$

$\Rightarrow 33275=5^2\times 7^3$

Hence, $33275$ must be divided by $25$ to make it a perfect cube.

$\therefore \frac{33275}{25}=1331={11}^3$

Consider $120393$,

$\begin{array}{cc}3& 120393\\ 3& 40131\\ 3& 13377\\ 7& 4459\\ 7& 637\\ 7& 91\\ 13& 13\\ & 1\end{array}$

$\Rightarrow 120393=3^3\times 7^3\times 13$

Hence, $120393$ must be divided by $13$ to make it a perfect cube.

$\therefore \frac{120393}{13}=9261=3^3\times 7^3={21}^3$