Find the roots of this 6th degree polynomial Hey guys I'm reviewing for a test and I'm getting stuck on one part, I can't remember what to do next.
$x^6+16x^3+64$
$(x^3)^2+16x^3+64$   
let $x^3=w$
$w^2+16w+64$
$(w+8)^2$ now substitute again
$(x^3+8)^2$
Now what do I do?

Then you have to solve $x^3+8=0$:
\begin{align*}
x^3+8=0\Longleftrightarrow
x^3=-8
\end{align*} 
But
$$
-8=8(-1)=8e^{i\pi}
$$
Thus
$$
x=2e^{ki\pi/3}\;\;\;\;k=0,1,2.
$$
Hence
$$
x^6+16x^3+64=(x^3+8)^3=(x-2)^3(x-2e^{i\pi/3})^3(x-2e^{2i\pi/3})^3
$$
