Find roots of $p(x)=x^3-6x-6$ using cubic formula
Quadratic Formula
$$ x^3 +ax^2 +bx+c=0$$
sub in $x=y-a/3$ to get
$$ y^3+py+1$$
where $$ \begin{aligned} p&=\frac{-a^2}{3}+b \\ q&= \frac{2a^3}{27}-\frac{ab}{3}+c \end{aligned}$$
$$ \begin{aligned} R&=\frac{1}{2}\left (-q + \sqrt{q^2+\frac{4 p^3}{27}}\right) \\ S&= \frac{1}{2} \left ( -q -\sqrt{q^2 +4p^3/27}\right) \end{aligned} $$
and solutions of $z$ are $$ z=R^{1/3}.R^{1/3} w ,R^{1/3} w^2 $$ and $$ z= S^{1/3},S^{1/3} w ,S^{1/3} w^2 $$
In this case so $a=0, b=-6,c=-6$ and $p=-6,q=-6$
getting that $R=-2$ and $S=-4$
From R
$$ z=(-2)^{1/3},(-2)^{1/3} w^1 , (-2)^{1/3} w^2$$
From S $$ z=(-4)^{1/3},(-4)^{1/3} w^1 , (-4)^{1/3} w^2$$
guessing that $w$ is the root of unite for 3. I want to verify with Sage that these are indeed roots
I am not gettting that $(-2)^{1/3}$ is a root. Howell has good section on this about tartaglia , Ferro . on his intro on later sections. But dont have the book with me
You didn't define $z$ above.
$x^3 - 6x - 6 = 0$
let $x = z + \frac {2}{z}$
$z^3 + 6z + \frac {12}{z} + \frac {8}{z^3} - 6z - \frac {12}{z} - 6 = 0\\ z^3 - 6 + \frac {8}{z^3} = 0\\ z^6 - 6z^3 + 8 = 0\\ (z^3 - 4)(z^3 - 2) = 0$
if $z = 4^{\frac13}$ then $\frac {2}{z} = 2^{\frac 13}$
$x = 4^{\frac13} + 2^{\frac 13}$
i.e. you need both $R$ and $S$ above to find $x$... and you flipped your signs.
That takes care of the real root.
Then $\omega(4^{\frac13} + 2^{\frac 13})$ and $\omega^2(4^{\frac13} + 2^{\frac 13})$
Will be your complex roots.
And yes $\omega, \omega^2$ are the roots of $(x^2 + x + 1)$ i.e. roots of unity.