<s>[INST] What is complex number z if $z^8+16z^4+256=0$? So far, I have set y to equal to $z^4$ and used the quadratic equation to solve $y = -8+8\sqrt{3}i$ or $-8-8\sqrt{3}i$. How do I determine the 8 different values of $z$?
 [/INST] $$z^8+16z^4+256=0$$
$$(\frac{z}{2})^8+(\frac{z}{2})^4+1=0$$
let $x=\frac{z}{2}$
$$x^8+x^4+1=0$$
$$(x^4+1)^2-x^4=0$$
$$(x^4-x^2+1)(x^4+x^2+1)=0$$
$$((x^2+1)^2-3x^2)((x^2+1)^2-x^2)=0$$
so
$$(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)(x^2-x+1)(x^2+x+1)=0$$
then you can use the quadratic formula  to find all the complex roots
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