Question

Is the following number the perfect cube? If yes, then find the number whose cube is the given number: \[1728\] .

Answer 1

Hint: In this problem we are required to find whether the given number \[1728\] is a perfect cube or not and the number whose cube is \[1728\] i.e., the cube root of this number. To solve this problem, we will use the Prime Factorisation Method. According to this method we will first find the prime factors of the given number and then we pair similar numbers in a group of three to denote them as cubes. From that we get the required given.

Complete step by step answer:
The number we are given is \[1728\] . Now, we have to find if this number is a perfect cube or not and if it is, then we must find the number whose cube is \[1728\] i.e., the cube root of \[1728\] .
To solve this problem, we will use the Prime Factorisation Method.
According to the Prime Factorisation Method we first write the prime factors of \[1728\] as
$\begin{align}
  & 2\left| \!{\underline {\,
  1728 \,}} \right. \\
 & 2\left| \!{\underline {\,
  864 \,}} \right. \\
 & 2\left| \!{\underline {\,
  432 \,}} \right. \\
 & 2\left| \!{\underline {\,
  216 \,}} \right. \\
 & 2\left| \!{\underline {\,
  108 \,}} \right. \\
 & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & ~~~3 \\
\end{align}$
 $\Rightarrow 1728=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$
Now, we pair the similar numbers from the above set of prime factors in a group of three as
$\Rightarrow 1728=\left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 3\times 3\times 3 \right)$
$\Rightarrow 1728={{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}$
Therefore, the number \[1728\] is a perfect cube.
Now, we apply cube root on both the left- and right-hand side of the above equation as shown below
\[\Rightarrow {{1728}^{\dfrac{1}{3}}}={{\left( {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}} \right)}^{\dfrac{1}{3}}}\]
Simplifying it further we get
\[\Rightarrow \sqrt[3]{1728}=\sqrt[3]{\left( {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}} \right)}\]
\[\Rightarrow \sqrt[3]{1728}=2\times 2\times 3\]
\[\Rightarrow \sqrt[3]{1728}=12\]

Therefore, the number \[1728\] is a perfect cube and the required given is $12$

Note: Instead of finding the cube root using the Prime Factorisation Method we can also use the Estimation Method. Here, it is necessary for students to remember the cubes of natural numbers $1$ to $9$ . Considering the unit digit of the given number and applying some other estimations we can find the cube root of a perfect cube.