Question

Which of the following is a perfect square as well as a cube 343, 125, 81 or 64?
A. 343
B. 125
C. 81
D. 64

Answer 1

Hint: We first find the prime factorisation of the given numbers. We take the indices form of the prime number and try to form the cue or square of them. We find the number which is both a cube and a square number.

Complete step by step answer:
We take all the four given terms 343, 125, 81 or 64 and check if they are perfect squares as well as a cube or not. We need to find the prime factorisation of the given number 343.
\[\begin{align}
  & 7\left| \!{\underline {\,
  343 \,}} \right. \\
 & 7\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Therefore, \[343=7\times 7\times 7={{7}^{3}}\]. Therefore, 343 is a cube number not a square number.

We need to find the prime factorisation of the given number 125.
$\begin{align}
  & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[125=5\times 5\times 5={{5}^{3}}\]. Therefore, 125 is a cube number not a square number.

We need to find the prime factorisation of the given number 81.
$\begin{align}
  & 3\left| \!{\underline {\,
  81 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[81=3\times 3\times 3\times 3={{3}^{4}}={{9}^{2}}\].Therefore, 81 is a square number not a cube number.

We need to find the prime factorisation of the given number 64.
\[\begin{align}
  & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Therefore, \[64=2\times 2\times 2\times 2\times 2\times 2={{2}^{6}}={{8}^{2}}={{4}^{3}}\].Therefore, 64 is both a square and a cube number.

Hence, the correct option is D.

Note: If we are finding square and cube roots of any numbers, we don’t always need to find the all-possible roots. The problem gets more complicated in that case. The simplification of the indices is the main criteria we need to find the simplified form of the given fraction.