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 

Putting $x=\cos(\frac{2\pi}{7})$ you have the polynomial $1+x-4x^2-8x^3\not= 0$ then $8x^3+4x^2-x-1\not=0$ from that
$$8x^3+4x^2-x-1=4x^2(2x+1)-x-1-x+x=4x^2(2x+1)-(2x+1)+x=(2x+1)(4x^2-1)+x=(2x+1)^2(2x-1)+x$$
Since $(2x+1)^2>0$ and $x>\frac{1}{2}$ since $\frac{2\pi}{7}<\frac{\pi}{3}$ we have that $(2x-1)>0$ and a sum of positive numbers isn't zero.
