Help please on complex polynomials I wanted to know if there's any good approaches to these questions
a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$
as a product of two real factors one of which is a quadratic.
b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z^6+z^3+1=0$
c)By letting $y=z+\frac{1}{z}$ and dividing $z^6+z^3+1=0$ by $z^3$, deduce that:
   $cos\frac{2\pi}{9}+cos\frac{4\pi}{9}+cos\frac{8\pi}{9}=0$
Any help would be greatly appreciated.

Hint: $a^3-b^3 = (a-b) (a^2+a b+b^2)$.  Think about this with $a=z^3$ and $b=1$.
