Factorise $16{{x}^{5}}-144{{x}^{3}}$
A. $16{{x}^{3}}\left( x-3 \right)\left( x-3 \right)$
B. ${{x}^{3}}\left( 2x+3 \right)\left( 2x-3 \right)$
C. ${{x}^{3}}\left( x+2 \right)\left( x-3 \right)$
D. $16{{x}^{3}}\left( x+3 \right)\left( x-3 \right)$

Hint: For this problem we need to factorize the given expression. For this we will first consider the coefficients in the given expression and write them in the prime factorization form. After that we will consider each and every variable and simplify them by using the exponential formula $a\times a\times a\times a\times a....\text{ n times}={{a}^{n}}$. After simplifying all the coefficients and variables, substitute them in the given expression and take all possible terms as common and simplify the equation to get the required result.

Complete step by step solution:
Given expression $16{{x}^{5}}-144{{x}^{3}}$.
In the given expression we have $16$, $144$ as coefficients and ${{x}^{5}}$, ${{x}^{3}}$ as variables.
Considering the coefficient $16$. The prime factorization of the number is given by
$\begin{align}
  & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
From this we can write the number $16$ as $16=2\times 2\times 2\times 2$.
Considering the coefficient $144$, the prime factorization of the number is given by
$\begin{align}
  & 2\left| \!{\underline {\,
  144 \,}} \right. \\
 & 2\left| \!{\underline {\,
  72 \,}} \right. \\
 & 2\left| \!{\underline {\,
  36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
From this we can write the number $144$ as $144=2\times 2\times 2\times 2\times 3\times 3$.
Considering the variable ${{x}^{5}}$ and writing the variable by using the exponential formula $a\times a\times a\times a\times a....\text{ n times}={{a}^{n}}$, then we will get
${{x}^{5}}=x\times x\times x\times x\times x$
Considering the variable ${{x}^{3}}$, by using the exponential form $a\times a\times a\times a\times a....\text{ n times}={{a}^{n}}$ we can write ${{x}^{3}}$ as
${{x}^{3}}=x\times x\times x$
Substituting all the values in the given expression $16{{x}^{5}}-144{{x}^{3}}$, then we will get
$16{{x}^{5}}-144{{x}^{3}}=2\times 2\times 2\times 2\left( x\times x\times x\times x\times x \right)-2\times 2\times 2\times 2\times 3\times 3\left( x\times x\times x \right)$
From the above equation we can take the term $2\times 2\times 2\times 2\left( x\times x\times x \right)$ as common, then we will get
$16{{x}^{5}}-144{{x}^{3}}=2\times 2\times 2\times 2\left( x\times x\times x \right)\left[ \left( x\times x \right)-3\times 3 \right]$
Simplifying the above equation by using the formula $a\times a\times a\times a\times a....\text{ n times}={{a}^{n}}$, then we will have
$16{{x}^{5}}-144{{x}^{3}}=16{{x}^{3}}\left( {{x}^{2}}-{{3}^{2}} \right)$
We have an algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Applying this formula in the above equation, then we will get
$16{{x}^{5}}-144{{x}^{3}}=16{{x}^{3}}\left( x+3 \right)\left( x-3 \right)$

So, the correct answer is “Option D”.

Note: We can also check whether the obtained answer is correct or wrong by multiplying the factors we got in the solution and comparing it with the given expression. If the both are the same then our answer is correct otherwise the calculated values are not the factors of the given expression.