What is the smallest number by which 8640 must be divided so that the quotient is a perfect cube?

Hint: This question can be done easily by prime factorization method
In order to find the smallest number to be divided first we factorize 8640.
$
   \Rightarrow 8640 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \\
\Rightarrow 8640 = {2^6} \times {3^3} \times 5 \\ $
In the above factorization, we find that there is a triplet of 2 and 3 but there is no triplet of 5.
Hence, 5 is the smallest number which must divide 8640 so that the quotient is a perfect cube.

Note: Any number which is a perfect cube will be multiple of a triplet of digits. Cube roots of the number can also be found out by the above mentioned prime factorization method.