 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. ThusAll six have modulus one, and their distinct arguments show that no roots have been repeated or omitted from the degree-six polynomial.