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Find all solutions to $z^6+z^4+z^3+z^2+1=0$

I attempted to simplify the left side as a sum of geometric progression that's missing the 5th and 1st power term $$\frac{z^7-1}{z-1}-(z^5+z)=0$$

(stuck at this point)

Am I on the right track? If so, how do you solve this equation? If not, what's the correct way of finding all roots of this equation using Roots of Unity?

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    $\begingroup$ This is a cubic equation in $w=z+z^{-1}$. $\endgroup$ – Lord Shark the Unknown Jan 5 '18 at 7:15
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We have: $$z^6+z^4+z^3+z^2+1$$ $$= z^6 + z^5 + z^4 + z^3 + z^2 - z^5 - z^4 - z^3-z^2-z + z^4 + z^3 + z^2+z+1 $$ $$= (z^2-z+1)(z^4+z^3+ z^2+z+1)=0$$

Surely you can take it from here.

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Hint:  $z=0\,$ is not a root, then dividing by $\,z^3\ne0\,$ gives: $\displaystyle z^3+\frac{1}{z^3}+z+\frac{1}{z} +1=0\,$.

Let $\displaystyle\,u=z+\frac{1}{z}\,$ so that $\displaystyle\,u^3 = z^3+\frac{1}{z^3} + 3\left(z + \frac{1}{z}\right) \iff z^3+\frac{1}{z^3} = u^3 - 3u\,$, then factor the resulting $\,u^3 - 3u +u+1 = 0 \iff (u-1)(u^2+u-1)=0\,$.

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