<s>[INST] Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution. 
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
 [/INST] Hint:
Since $$z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)$$
Solve $z^2+z+1=0$ and $z^2-z+1=0$ by using the quadratic formula:
$$z^2+z+1=0\implies z_{1,2}=\frac{-1\pm\sqrt{1-4}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
$$z^2-z+1=0\implies z_{3,4}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
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