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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.

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.