#### Answer

$x=\left\{ -1,0,1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
x^3-x=0
,$ in factored form. Then equate each factor to zero (Zero Product Property). Finally, solve each equation.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=x$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x(x^2-1)=0
.\end{array}
The expressions $
x^2
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^2-1
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x[(x)^2-(1)^2]=0
\\\\=
x[(x+1)(x-1)]=0
\\\\=
x(x+1)(x-1)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions to the equation above are
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+1=0
\\\\\text{OR}\\\\
x-1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+1=0
\\\\
x=-1
\\\\\text{OR}\\\\
x-1=0
\\\\
x=1
.\end{array}
Hence, $
x=\left\{ -1,0,1 \right\}
.$