= Solution
Set $w=z^3$. Then $w^2+w+1=0$ gives $w=e^{2\pi i/3}$ or $e^{-2\pi i/3}$. Taking all three cube roots of each yields six distinct solutions.
Equivalently, multiply the original <polynomial> by $z^3-1$ to obtain $z^9-1$. The original roots are exactly the ninth <roots of unity> that are not cube <roots of unity>: a cube root would give $1+1+1=3$, not zero. Thus
$$
\boxed{z=e^{2\pi i k/9},\qquad k\in\{1,2,4,5,7,8\}.}
$$
All six have modulus one, and their distinct arguments show that no roots have been repeated or omitted from the degree-six <polynomial>.
Back to article page
