How do you factor $x^3 - 8 = 0$? I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$? 

Let $x = y-1$.
Then $x^3 - 8 = y^3 - 3y^2 + 3y - 9 = (y-3)(y^2 + 3)$.
Setting this last expression equal to zero, we find all three roots:
One is $y = 3$ and the other two are $y = \pm i\sqrt{3}$.
Since $x = y-1$, we conclude the roots are $x = 2$ and $x = \pm i\sqrt{3} - 1$.
"QED"
