#### Answer

$x=0,\pm 1$

#### Work Step by Step

$f(x)=x^3-x$
Since we are solving for all the real values of $x$ when $f(x) = 0$, set the function $f(x)=x^3-x$ equal to zero and we get the following equation:
$x^3-x=0$
First, factor an $x$ out of the left side of the equation.
$x(x^2-1)=0$
Next, you can see that the above equation holds true when $x=0$ or $x^2-1=0$. Since $x=0$ is already solved, we know that is a solution. Yet, we still need to solve for the second equation as shown below:
$x^2-1=0$
Add 1 to both sides.
$x^2-1+1=0+1$
$x^2=1$
Now solve for x by taking the square root of both sides.
$\sqrt{x^2}=\sqrt{1}$
Remember that taking the square root of both sides means we need to include the $\pm$ on the right side of the equation.
$x=\pm 1$
So, the final solution is $x=0,\pm 1$