Question

Is $9720$ a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

Answer 1

Step 1. Find the prime factor of the given number :

The prime factors of $9720$ are

$9720=2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 5$

Step 2. Find the perfect cube root :

On grouping and pairing the prime factors we get

$$\begin{gathered} 9720=\left(2\times 2\times 2\right)\times \left(3\times 3\times 3\right)\times 3\times 3\times 5 \\ \Rightarrow 9720=2^3\times 3^3\times 3\times 3\times 5 \end{gathered}$$

Factor $3$ and $5$ has no pair

Therefore $9720$ is not a perfect cube

Step 3. Find the smallest number :

On grouping and pairing the prime factors we get ,

$$\begin{gathered} 9720=\left(2\times 2\times 2\right)\times \left(3\times 3\times 3\right)\times 3\times 3\times 5 \\ \Rightarrow 9720=2^3\times 3^3\times 3\times 3\times 5 \end{gathered}$$

Two $3$ doesn't form any pair and one $5$ doesn't form any pair

Therefore $3\times 3\times 5=45$ must be divided to get perfect cube

Hence, $9720$ is not a perfect cube and $45$ should be divided to get perfect cube.