How do you factor ${{x}^{3}}-9x=0$?

Hint: The given equation in the above question is cubic. This means that it must be factored into three monomials. For this, we need to take the factor $x$ common from both the terms on the left hand side of the given equation. Then, on applying the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ on the quadratic factor hence obtained, we can completely factorise the left hand side, and hence the given equation.

Complete step by step solution:
The equation is written in the above question as
$\Rightarrow {{x}^{3}}-9x=0$
Now, we can see that the factor $x$ is common to both the terms on the left hand side of the above equation. Therefore taking $x$ common on the LHS, we get
$\Rightarrow x\left( {{x}^{2}}-9 \right)=0$
We know that nine is equal to the square of three. So we can substitute $9={{3}^{2}}$ in the above equation to rewrite it as
$\Rightarrow x\left( {{x}^{2}}-{{3}^{2}} \right)=0$
Now, we know the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Substituting $a=x$ and $b=3$ in this identity, we get
\[\Rightarrow {{x}^{2}}-{{3}^{2}}=\left( x+3 \right)\left( x-3 \right)\]
Substituting this in the above equation, we finally obtain
$\Rightarrow x\left( x+3 \right)\left( x-3 \right)=0$
Hence, the given equation is completely factored.

Note: There can be many methods of factoring the given equation. For example, we can also factorize the given equation using the hit and trial to guess its root. Let us suppose that we got the guessed root as $3$, then by factor theorem we can say that $\left( x-3 \right)$ will be the factor of the polynomial on the LHS. Then on dividing the LHS by $\left( x-3 \right)$ we will get factorised form of the given equation.