I proved that $z^6+z^3+1 = (z^2-2z\cos(\frac{2π}{9})+1)(z^2-2z\cos(\frac{4π}{9})+1)(z^2-2z\cos(\frac{8π}{9})+1)$
Using this, how to show that $$2\cos(3 \theta)+1=8\left(\cos \theta-\cos\left(\frac{2π}{9}\right)\right)\left(\cos \theta-\cos\left(\frac{4π}{9}\right)\right)\left(\cos\theta-\cos\left(\frac{8π}{9}\right)\right)$$
I know it involves some trig substitution but I'm not sure how. Help appreciated, thanks.
Divide by $z^3$: $$z^3+z^{-3}+1=(z+z^{-1}-2\cos(2\pi/9))(z+z^{-1}-2\cos(4\pi/9))(z+z^{-1}-2\cos(8\pi/9))$$ Put in $z=\exp(i\phi)$, noting that $z+z^{-1}=2\cos\phi$ and $z^3+z^{-3}=2\cos3\phi$: $$2\cos3\phi+1=(2\cos\phi-2\cos(2\pi/9))(2\cos\phi-2\cos(4\pi/9))(2\cos\phi-2\cos(8\pi/9)).$$