# Complex roots of $z^6 + z^3 + 1 = 0$

The equation I'm trying to solve is $f(z) = 0$ where

$$f(z) = z^6 + z^3 + 1$$

I already tried the following: randomly throwing in complex numbers and real numbers, rational root theorem, banging my head on the table, and other painful things. Any ideas where I can start?

Following Potato's hint, we have that if we set $y = z^3$ the resulting quadratic would be $y^2 + y + 1 = 0$, by which we can use the quadratic formula, so that

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

And then take the cube roots of the solutions. Thanks guys.

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Set $y=z^3$, solve the resulting quadratic in $y$, the take cube roots of those solutions. – Potato Mar 11 '13 at 1:11
@Potato Thanks, I'll edit my question and put the answer in. – noobProgrammer Mar 11 '13 at 1:14

Hint: $(x^6+x^3+1)(x^3-1)=x^9-1$