Factorize $z^6 + 1$, and deduce $\cos{3x} = \cos{x}(2\cos{x}+\sqrt{3})(2\cos{x}-\sqrt{3})$

Question

Factorise $z^6 + 1$ into real quadratic factors.

Hence, deduce that $\cos{3x} = \cos{x}(2\cos{x}+\sqrt{3})(2\cos{x}-\sqrt{3})$

I factorised $z^6+1$ into $(z^2+1)(z^4-z^2+1)$, then tried using De Moivre to get complex roots, but I ended up nowhere.

Answer 1

$$z^6+1$$ $$=(z^2+1)(z^2+\sqrt 3z+1)(z^2-\sqrt 3z+1)$$ $$=(z^4+z^2)(z+\sqrt 3+1/z)(z-\sqrt 3+1/z)$$ Divide by $z^3$: $$z^3+z^{-3}=(z+1/z)(z+\sqrt 3+1/z)(z-\sqrt 3+1/z)$$ Let $z=e^{ix}$, then $z+1/z=2\cos x$ and $z^3+z^{-3}=2\cos 3x$. $$2\cos 3x=2\cos x(2\cos x+\sqrt 3)(2\cos x-\sqrt 3)$$ $$\cos 3x=\cos x(2\cos x+\sqrt 3)(2\cos x-\sqrt 3)$$