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If $x = {2^3} \times {4^2} \times {\left( {17} \right)^3}$, then which number should be divided by $x$ to get a perfect cube.
(A) 2
(B) 4
(C) 8
(D) 17

Answer
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Hint- Here, we will proceed by representing the given number as the product of powers of prime factors and then comparing it to the general form of any number which is a perfect cube.
Given, $x = {2^3} \times {4^2} \times {\left( {17} \right)^3}$
Let us represent the given number $x$ as the product of prime factors.
$x = {2^3} \times {4^2} \times {\left( {17} \right)^3} = {2^3} \times {\left( {{2^2}} \right)^2} \times {\left( {17} \right)^3} = {2^3} \times {\left( 2 \right)^4} \times {\left( {17} \right)^3} = {\left( 2 \right)^7} \times {\left( {17} \right)^3}$
As we know that for a number to be a perfect cube, it should be of the form ${\left( a \right)^{3b}} \times {\left( c \right)^{3d}} \times {\left( e \right)^{3f}}$ where $a$, $c$ and $e$ are the prime factors of the given number and $b$, $d$ and $f$ are any numbers i.e., the power of the prime factors should be a multiple of 3.
Clearly, the given number $x$ can be represented in the above form as under
\[
  x = {\left( 2 \right)^7} \times {\left( {17} \right)^3} = {\left( 2 \right)^{6 + 1}} \times {\left( {17} \right)^3} = {\left( 2 \right)^6} \times 2 \times {\left( {17} \right)^3} \Rightarrow \dfrac{x}{2} = {\left( 2 \right)^6} \times {\left( {17} \right)^3} \\
   \Rightarrow \dfrac{x}{2} = {\left[ {{2^2} \times 17} \right]^3} \\
 \]
In the above equation, \[{\left[ {{2^2} \times 17} \right]^3}\]is a perfect cube which is equal to \[\dfrac{x}{2}\] so the given number $x = {2^3} \times {4^2} \times {\left( {17} \right)^3}$ should be divided by 2 in order to get a perfect cube.
Therefore, option A is correct.

Note- When bases are same then powers are added in multiplication i.e., ${a^c} \times {a^d} \times {a^e} = {\left( a \right)^{c + d + e}}$ and when bases are same then powers are subtracted in division i.e., $\dfrac{{{a^c}}}{{{a^b}}} = {\left( a \right)^{c - b}}$.