You may want to review solving
quadratic equations in *x*^{2} first.

The same thing works if you put *x*^{3} in place
of *x*^{2}. For example, if you have to solve
8*x*^{6} + 6*x*^{3} + 1 = 0, you can use
factoring (or the Quadratic Formula) to work out
that *x*^{3} = −1 ⁄ 2 or
−1 ⁄ 4. Then *x* =
−1 ⁄ 2 * ^{3}√4 or
−1 ⁄ 2 * ^{3}√2.

There's a twist, though. If we allow complex numbers, then
numbers will have 3 cube roots, not just one. So there are
actually 4 more possibilities for *x*. They're not real,
though, so we ignore them.