You may want to review solving quadratic equations in x2 first.
The same thing works if you put x3 in place of x2. For example, if you have to solve 8x6 + 6x3 + 1 = 0, you can use factoring (or the Quadratic Formula) to work out that x3 = −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.