Find the smallest numbers by which $968$ should be multiplied to make it a perfect cube.

Hint: For a number to be a perfect cube , its prime factors must occur in triplets. We first find the prime factor; Prime factorization is a way to write a composite number as the product of prime factors. Prime factors are those numbers or factors which are greater than one and have exactly two factors one is the number itself and other is one.

Complete step by step solution:
The given number is $968$ , Finding prime factors,we know that the number $2$ is the smallest prime number. So, in order to find prime factors of $968$ , let us first divide $968$ by least prime number i.e., $2$.
$968 \div 2 = 484$
Again,
$484 \div 2 = 242$
Again,
$242 \div 2 = 121$
Now, we know that $121$ is not divisible by $2$ so we move to the next prime number known to us, which is $3$. But it is not divisible by this also. So we move to $5$, still not divisible. So, we moved to $7$. Still not divisible. So, we moved to $11$.
$121 \div 11 = 11$
Again,
$11 \div 11 = 1$ .
$\therefore 968 = 2 \times 2 \times 2 \times 11 \times 11$
We can see that $11$ does not occur in triplets. Therefore, $11$ should be multiplied with the given number that is $968$ to make it a perfect cube.

Hence the number is $11$.

Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. There is another way of calculating prime factors of numbers , which is by factor tree. Under this way, we split the number into its prime factors.