How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:

$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$

Wolframalpha shows this, but I can't prove it.
http://www.wolframalpha.com/input/?i=1%2Bcos(2pi%2F7)-4cos%5E2(2pi%2F7)-8cos%5E3(2pi%2F7)%3D0 

We prove a closely related result, which in particular shows there is a typo in the given equation.
The number $e^{2\pi i/7}$ is a root of $x^7=1$, and therefore of $x^6+x^5+\cdots+x+1=0$, or equivalently of
$$(x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=0.\tag{1}$$
(We divided through by $x^3$.) Let $w=\frac{1}{2}(x+x^{-1})$. 
Note that $x^3 +x^{-3}=8w^3-6w$ and $x^2+x^{-2}=4w^2-2$ and $x+x^{-1}=2w$ So our equation can be rewritten as
$$8w^3+4w^2-4w-1=0.\tag{2}$$
Since $e^{2\pi i/7}$ is a root of (1), it follows that $\cos(2\pi/7)$ is a root of (2).
