[Math] Complex roots of $z^6 + z^3 + 1 = 0$

Question

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?

EDIT: Answer

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.

Answer 1

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