<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] $$\begin{align}z^6&=1\\
z^6-1&=0\\
(z-1)(\underbrace{z^5+z^4+z^3+z^2+z+1}_{f(z)})&=0
\end{align}$$
Hence the solutions to $f(z)$ are
$$\begin{align}z^6&=1=e^{i2\pi} &&(z\neq1)\\
z&=e^{i\frac {2n\pi}6}=e^{i\frac {n\pi}3}&& (n=1,2,3,4,5)\quad\blacksquare\end{align}$$
</s>