How do you factor the expression ${{x}^{4}}+216x$?

Hint: For this problem we need to calculate the factors of the given equation. We can observe that the given equation has a common term $x$ in the whole equation. So, we will first take $x$ as common from the whole equation. Now we will get the sum of the two cubes in the equation. We have the algebraic formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. so, we will apply this formula and simplify the equation to get the required result.

Complete step by step solution:
Given that, ${{x}^{4}}+216x$.
Taking $x$ as common from the whole equation, then we will get
$\Rightarrow {{x}^{4}}+216x=x\left( {{x}^{3}}+216 \right)$
Considering the term ${{x}^{3}}+216$. The first term which is ${{x}^{3}}$ already in cubic form. So, we need to write the second term $216$ as cube value. When we calculated the prime factors of value $216$, we can write $216={{2}^{3}}\times {{3}^{3}}={{6}^{3}}$. Substituting this value in the above equation, then we will have
$\Rightarrow {{x}^{4}}+216x=x\left( {{x}^{3}}+{{6}^{3}} \right)$
In the above equation we can observe the sum of two cubes which is in the form of ${{a}^{3}}+{{b}^{3}}$, where $a=x$, $b=6$. In algebra we have the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. Using this formula and substituting the known values in the above equation, then we will get
$\begin{align}
  & \Rightarrow {{x}^{4}}+216x=x\left( x+6 \right)\left( {{x}^{2}}-\left( x \right)\left( 6 \right)+{{6}^{2}} \right) \\
 & \Rightarrow {{x}^{4}}+216x=x\left( x+6 \right)\left( {{x}^{2}}-6x+36 \right) \\
\end{align}$
Hence the factors of the given equation ${{x}^{4}}+216x$ are $x$, $x+6$, ${{x}^{2}}-6x+36$.

Note: We can also check whether the obtained result is correct or wrong. When we multiplied all the factors, we calculated then we should get the given equation. Then only our calculated answer is correct otherwise our solution is incorrect.