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**Hint:**We try to form the indices formula for the value 3. This is finding the cube root of 9720. We find the prime factorisation of 9720. Then we take one digit out of the three same number of primes. There will be an odd number of primes remaining in the root which can’t be taken out.

**Complete step by step answer:**

We try to find the value of the algebraic form of $\sqrt[3]{9720}$. This is a cube root form.

We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 3 we get \[{{a}^{\dfrac{1}{3}}}=\sqrt[3]{a}\].

We need to find the prime factorisation of the given number 9720.

$\begin{align}

& 2\left| \!{\underline {\,

9720 \,}} \right. \\

& 2\left| \!{\underline {\,

4860 \,}} \right. \\

& 2\left| \!{\underline {\,

2430 \,}} \right. \\

& 3\left| \!{\underline {\,

1215 \,}} \right. \\

& 3\left| \!{\underline {\,

405 \,}} \right. \\

& 3\left| \!{\underline {\,

135 \,}} \right. \\

& 3\left| \!{\underline {\,

45 \,}} \right. \\

& 3\left| \!{\underline {\,

15 \,}} \right. \\

& 5\left| \!{\underline {\,

5 \,}} \right. \\

& 1\left| \!{\underline {\,

1 \,}} \right. \\

\end{align}$

Therefore, \[9720=2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 5\].

For finding the cube root, we need to take one digit out of the three same number of primes.

This means in the cube root value of 9720, we will take out one 2 and one 3 from the multiplication. But there will still remain two 3s and one 5 which can’t be taken out.

Therefore, to make it a perfect cube we divide 9720 with the least number of \[3\times 3\times 5=45\].

9720 is not a perfect cube. The smallest number by which it should be divided to get a perfect cube is 45.

**Note:**We can also use the variable form where we can take $x=\sqrt[3]{9720}$. But we need to remember that we can’t use the cube on both sides of the equation $x=\sqrt[3]{9720}$ as in that case we are taking two extra values as a root value. Then this linear equation becomes a cubic equation.

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