Find the smallest number which should be multiplied to \[432\] to get a perfect cube.

Hint: Perfect cube is a number which has all its prime factors as a group of $3$ numbers.

Complete step by step answer:
Given number is \[432\]
Writing \[432\] in prime factor from

(2) Therefore, prime factors of \[432\]are \[2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 2\]\[ = {2^4} \times {3^3}\]
(3) Writing prime factor in group of three \[ = {2^4} \times {3^3}\]
(4) In the above step, we see that $2$ is the only number which is not in the group of three.
Hence, we can say that \[432\] is not a perfect cube number.
(5) Now to make \[432\] as a perfect cube number, we should multiply it with such a number that it completes the three groups of elements of $2$. (i.e. by $4$)
(6) Therefore, on multiplying \[432\] by $4$, we get prime factors as
\[ = {2^6} \times {3^3}\]
\[ = {2^3} \times {2^3} \times {3^3}\]
$ = {\left( {2 \times 2 \times 3} \right)^3}$
$ = {\left( {12} \right)^3}$
$ = 12 \times 12 \times 12$
$ = 1728$
We see that all above prime factors are in a group of three elements. Hence, the number is a perfect cube.
(7) So, $4$ is the least number by which \[432\] is to be multiplied to make it a perfect cube. $1728$ is a perfect cube.

Note: Perfect cube is a number having all factors in a group of triplets.