Question

Is $729$ a perfect cube?

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 will also find that if it is a perfect cube, then whose perfect cube it is. First, find the prime factors of the number and then, 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 the 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 - $729$. 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 $729$.
\[
  3\left| \!{\underline {\,
  {729} \,}} \right. \\
  3\left| \!{\underline {\,
  {243} \,}} \right. \\
  3\left| \!{\underline {\,
  {81} \,}} \right. \\
  3\left| \!{\underline {\,
  {27} \,}} \right. \\
  3\left| \!{\underline {\,
  9 \,}} \right. \\
  3\left| \!{\underline {\,
  3 \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  1 \,}} \right. \\
 \]
Therefore, $243 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$.
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 $729$ 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 729 = (3 \times 3 \times 3) \times \left( {3 \times 3 \times 3} \right)$
As we can see that we could form two complete triplets and no prime factors are left out after grouping, $729$ is a perfect cube.
Next step is to find out whose cube $729$ is.
Now, we will take one number from each triplet and multiply them. The resultant product will be the answer.
$ \Rightarrow \sqrt[3]{{729}} = 3 \times 3 = 9$

Therefore, it is a cube of 9.

Note: We answered the question but what if the number was not a perfect cube, then how would you know it?
Let us assume the number $243$. If we find its prime factors and group together the triplets, we will get-
$ \Rightarrow 243 = (3 \times 3 \times 3) \times 3 \times 3$
As we can see that we could not group together all the factors into triplets and two factors still remain ungrouped, the number is not a perfect cube.