Quadratic Equation/Examples/z^6 + z^3 + 1 = 0/Proof 1

Proof
Although this is a sextic in $z$, it can be solved as a quadratic in $z^3$.

From the Quadratic Formula:

$-\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2$ are recognised as the complex cube roots of unity, and so:

It remains to solve for $z$.

By Roots of Complex Number:


 * $z = \set {w \alpha^k: k \in \set {1, 2, \ldots, n - 1} }$

where:
 * $w$ is a cube root of $e^{\pm 2 i \pi / 3}$
 * $\alpha$ is one of the complex cube roots of unity.

Taking $z^3 = e^{2 i \pi / 3}$:

The roots obtained by taking $z^3 = e^{-2 i \pi / 3}$ can be calculated as above:

We note that:

and: