find all complex solutions of $z^{12}-z^6-2z^3+2=0$ How could I find all the complex solutions to: $z^{12}-z^6-2z^3+2=0$
I tried substituting $y=z^3$ to get a simpler equation and I managed to get $y-1=0$ or $y^3+y^2-2=0$, but finding all complex solutions of the latter equation was also complicated.
 A: The equation $y^3+y^2-2$ has an obvious root: $y=1$, and it factorises as 
$y^3+y^2-2=(y-1)(y^2+2y+2)=(y-1)((y+1)^2+1)$, hence the initial equation is
$$y^4-y^2-2y+2=(y-1)^2((y+1)^2+1)=0$$
The solutions are 
$$y=1,\quad y=-1\pm\mathrm i.$$
There remains to compute the cubic roots of these numbers.
A: HINT:
$$z^{12}-z^6-2z^3+2=0\Longleftrightarrow$$
$$(z-1)^2(z^2+z+1)^2(z^6+2z^3+2)=0\Longleftrightarrow$$
$$(z-1)^2=0\Longleftrightarrow\space\space\vee\space\space(z^2+z+1)^2=0\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z-1=0\Longleftrightarrow\space\space\vee\space\space z^2+z+1=0\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z^2+z=-1\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z^2+z+\frac{1}{4}=-\frac{3}{4}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space \left(z+\frac{1}{2}\right)^2=-\frac{3}{4}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z+\frac{1}{2}=\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$

Substitute $x=z^3$:

$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space x^2+2x=-2\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space (x+1)^2=-1\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space x+1=\pm i\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space x=-1\pm i\Longleftrightarrow$$
$$z=1\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\space\space\vee\space\space z^3=-1\pm i$$
