How would one solve $z^6+z^4+z^3+z^2+1=0$? I recently came upon the following equation, which is supposed to be solvable using complex numbers:
$$z^6+z^4+z^3+z^2+1=0$$
The problem is that it is a degree $6$ polynomial with many other exponential terms, which I don't know how to deal with. I try substituting $w=z^2$ to try and make the equation simplier:$$w^3+w^2+wz+w+1=0$$but I can't make any progress with this either. Only thing that I see that could be used is that $w^3,w^2,w,1$ is a geometric series, not sure how that helps though.

Note that neither $z = 1$ nor $z = -1$ are roots. Then our equation becomes equivalent to
$$0 = (z^2 - 1)(z^6 + z^4 + z^3 + z^2 + 1) \text{ and } z \neq \pm 1.$$
This equation simplifies to
$$z^8 + z^5 - z^3 - 1 = 0.$$
We can factor this by:
$$z^8 + z^5 - z^3 - 1 = z^5(z^3 + 1) - 1(z^3 + 1) = (z^5 - 1)(x^3 + 1) = 0.$$
From this, and the fact that $z \neq \pm 1$, you can get the result easily.
How did I get this answer? It was mostly inspired by jlammy's answer, but one of the things that jumped out at me was that $z^3$ was an odd term out of a geometric series $z^6 + z^4 + z^2 + 1$. This meant that multiplying by $z^2 - 1$ would achieve some kind of simplification. I thought it was worth a shot.
In general, if you have a polynomial, whose coefficients are $\pm 1$, and you suspect that some roots of unity are involved, it's probably not a bad idea to try multiplying by $z^2 - 1$ anyway. It should take any of the cylotomic factors like $z^4 + z^3 + z^2 + z + 1$ or $z^2 - z + 1$, and turn them into simpler binomial factors like $z^5 - 1$ or $z^3 + 1$.
