${x}^{3}-16x=0$
Factor out $x$ from the expression$x \times \left( {x}^{2}-16 \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }x=0,\\{x}^{2}-16=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=0,\\x=-4,\\x=4\end{array}$
The equation has $3$ solutions$\begin{array} { l }x_1=-4,& x_2=0,& x_3=4\end{array}$