Question

Is $243$ a perfect cube? In case of a perfect cube, find the number whose cube it is.

Answer 1

Hint: In this question, we are given a number and we have been asked to find whether the given number is a perfect cube or not. In addition to this, we have also been asked if it is a perfect cube, then whose perfect cube it is. First, find the prime factors of the number and the, make groups of triplets of prime factors. If the triplets can be made, the number is a perfect cube, otherwise not. In case it is a perfect cube, multiply one number of the triplet with the other number of other triplet. Do this until all the triplets have been considered and you will get the required number.

Complete step-by-step solution:
We are given a number - $243$. We have to find out whether the number is a perfect cube or not.
We will begin by finding the prime factors of the number $243$.
\[\begin{array}{*{20}{c}}
  {{\text{ }}3\left| \!{\underline {\,
  {243} \,}} \right. } \\
  {{\text{ }}3\left| \!{\underline {\,
  {81} \,}} \right. } \\
  {{\text{ }}3\left| \!{\underline {\,
  {27} \,}} \right. } \\
  {3\left| \!{\underline {\,
  9 \,}} \right. } \\
  {3\left| \!{\underline {\,
  3 \,}} \right. } \\
  {{\text{ }}\left| \!{\underline {\,
  1 \,}} \right. }
\end{array}\]
Hence, $243 = 3 \times 3 \times 3 \times 3 \times 3$.
Next step is to group together the triplets of these prime factors. After that, we will see if any number is left out or not. If a number is left out, then $243$ is not a perfect cube. On the other hand, if we can make triplets out of all the prime factors, then the number is a perfect cube.
Grouping together the triplets,
$ \Rightarrow 243 = (3 \times 3 \times 3) \times 3 \times 3$
As we can see that we could form only one triplet and two prime factors are left out after grouping.

Since groups cannot be made, $243$ is not a perfect cube.

Note: We answered the question but what if the number was a perfect cube, then how would you find the number whose cube it is?
Let us assume the number $729$. If we find its prime factors and group together the triplets, we will get-
$ \Rightarrow 729 = (3 \times 3 \times 3) \times (3 \times 3 \times 3)$
As we can see that we could group together all the factors into triplets, the number is a perfect square. Let us find the number of whose cube $729$ is.
We will take one number from each triplet and multiply them with each other. The resultant product will be our answer. $\sqrt[3]{{729}} = 3 \times 3 = 9$
This is how we calculate cube roots.