How do you factor ${{x}^{3}}-81x$ ?

Hint: We first take $x$ common from the two terms of the expression. This simplifies the expression into $x\left( {{x}^{2}}-81 \right)$. We then factorize ${{x}^{2}}-81$ using the formula for the subtraction of two squares which is
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
So, $81$ being ${{9}^{2}}$ , ${{x}^{2}}-81$becomes as $\left( x+9 \right)\left( x-9 \right)$ . This way we reach our final answer.

Complete step by step answer:
The given expression is
${{x}^{3}}-81x$
As we can see that there is a $x$common in both the terms, so we take it common from these two terms. The expression thus becomes,
$\Rightarrow x\left( {{x}^{2}}-81 \right)$
$81$being a perfect square, can be written as ${{9}^{2}}$. The expression thus becomes,
$\Rightarrow x\left( {{x}^{2}}-{{9}^{2}} \right)....\text{expression}1$
The quadratic term in the above expression can be compared to the general expression ${{a}^{2}}-{{b}^{2}}$ which can be factorized as,
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)....\text{expression}2$
Comparing $\text{expression}1$ with $\text{expression}2$ we get,
$a=x$ and $b=9$
Factorising $\text{expression}1$accordingly, we get,
$\Rightarrow x\left( x+9 \right)\left( x-9 \right)$
Therefore, we can conclude that the given expression ${{x}^{3}}-81x$ can be factorized as $x\left( x+9 \right)\left( x-9 \right)$ .

Note: Students may get confused after seeing the cubic expression, so its better to rewrite it as the sum of a linear and a quadratic term by taking $x$ common. The given expression can also be solved in another way. If we observe closely, we can see that there are three values of $x$for which the expression turns out to be $0$. One of the values is $9$ . This means that $\left( x-9 \right)$ is a factor of the given expression. Dividing the given expression by $\left( x-9 \right)$ gives a quotient which gives us the other two factors. This process is nothing but the vanishing factor method which is often used.