What is the smallest number by which $392$ must be multiplied so that the product is a perfect cube?

The correct option is B: $7$

Given number is $392.$

Prime factorisation of $392$ is:
$\begin{array}{cc}2& 392\\ 2& 196\\ 2& 98\\ 7& 49\\ & 7\end{array}$

Hence, $392=2\times 2\times 2\times 7\times 7$

A number will a perfect cube, if all prime factors form a triplet.

Here, the number $7$ occurs only twice.

Clearly, $392$ must be multiplied by $7$ in order to make it a perfect cube.

$\Rightarrow 392\times 7=2\times 2\times 2\times 7\times 7\times 7$

$\Rightarrow 2744=2^3\times 7^3$

Hence, $7$ is the smallest number that should be multiplied to $392$ to give a perfect cube.