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 
 A: 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).
A: Here, we will prove the equation by @AndreNicolas and @Blue in a more elementary manner. 

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

Note that since $\cos 3x=4\cos^3 x-3\cos x$, and since $\cos 2x=2\cos^2 x-1$, our equation simplifies to $$  \cos \left(\frac{\pi}{7}\right) -\cos \left(\frac{2\pi}{7}\right) + \cos \left(\frac{3\pi}{7}\right) = \frac{1}{2}$$
Let $x=\frac{\pi}{7}$.
$$\cos x - \cos 2x + \cos 3x = \frac{1}{2}$$
$$\cos x + \cos 3x + \cos 5x = \frac{1}{2}$$
$$\cos x + \cos 3x + \cos 5x + ... = \frac{{\sin 2nx}}{{2\sin x}}$$
Since $n = 3$
$$\cos x + \cos 3x + \cos 5x = \frac{{\sin 6x}}{{2\sin x}}$$
$$\frac{{\sin 6x}}{{2\sin x}} = \frac{1}{2} \Leftrightarrow \sin 6x = \sin x$$
Which is true since $\sin (\pi-x)=\sin x$.
Or,similarly if $$K=\cos x - \cos 2x + \cos 3x $$ then $$K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}$$
Since $ 2\sin A\cos B=\sin(A+B)+\sin(A-B)$
A: 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.
A: I will address the problem as it has been stated not as to whether there is a conjectured or assumed typo. 
First off the problem is equivalent to showing
$$\frac{(1+\zeta)}{\zeta^{2}(1+2\zeta)}\neq4$$ where $\zeta=cos(2\pi/7)$. 
Since $1/2\lt\zeta\lt1$, the left hand side of the inequality is less than $3$ and therefore we are done.
A: Consider $p(x)=1+x-4x^2-8x^3$, that have only one real root, $x_0\approx0,4$. Now, calculating $x=\cos(2\pi/7)\approx0,6$ we find that $x\neq x_0$
A: We consider $$f(x) = x - 4x^2 - 8x^3  $$
Notice that for $x \ge \frac{1}{2}, \>\> f(x) < -1$ siince $f(\frac{1}{2}) = -\frac{3}{2} $ and $f'(\frac{1}{2}) < 0$ and $f''(x) < 0$ for all $x > -1$ 
Now it's only important to show that $\cos \frac{2 \pi}{7} \ge \frac{1}{2}$
It should be relatively easy to show $\cos \frac{2 \pi}{7} \ge \cos \frac{\pi}{3}$
