<s>[INST] 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?
 [/INST] We can actually factor anything in the form $x^3+a^3$ into $(x+a)(x^2-ax+a^2)$.
$x^3+8$ can be factored into the following:
$$x^3+8 = x^3+2^3 =(x+2)(x^2-2x+4)$$
We actually have $(x^3+8)^2$, however we are setting this value to equal $0$ to find the roots so we can just take the square root of both sides and obtain:
$$x^3+8 = 0$$
$$(x+2)(x^2-2x+4) = 0$$
We know now that $x+2 = 0$ so $-2$ is a solution. We also know that $x^2-2x+4=0$ gives us a solution. So if we use the quadratic formula, we get:
\begin{align*}
\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot1\cdot4}}{2\cdot1}&= \frac{2\pm\sqrt{-12}}{2} \\
&= \frac{2\pm2\sqrt{3}i}{2} \\
&= {1\pm\sqrt{3}i}
\end{align*}
So our solutions are:
$$\boxed{-2, {1+\sqrt{3}i}, {1-\sqrt{3}i}}$$
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