## Algebra and Trigonometry 10th Edition

$x=0,\pm 1$
$f(x)=x^3-x$ 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 $x=0,\pm 1$