Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$ 
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.

I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
 A: Put $2x=z+\dfrac1z$  in $$(2x)^3-(2x)^2-2(2x)+1=0$$
and multiply by $z+1$ to find $z^7+1=0$ whose roots are are $e^{(2k+1)\pi i/7}$ where $k\equiv0,\pm1,\pm2,\pm3\pmod7$
So, the roots of $$\dfrac{z^7+1}{z+1}=0$$ are $e^{(2k+1)\pi i/7}$ where $k\equiv0,\pm1,\pm2,3\pmod7$
Finally $2\cos y=e^{iy}+e^{-iy}$
A: $$     \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=     $$
$$  =\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}.  $$
Let $\cos\frac{\pi}{7}=x$. 
Hence,
$$2x^2-(4x^3-3x)-x=-\frac{1}{2},$$
which gives your equation.
A: We have
$$ \cos{2\theta} = 2\cos^2{\theta}-1 \\
\cos{3\theta} = \cos{\theta}(2\cos^2{\theta}-1) - 2\sin^2{\theta}\cos{\theta} = 4\cos^3{\theta}-3\cos{\theta}, $$
so
$$ \cos^2{\theta} = \frac{1}{2}(\cos{2\theta}+1) \\
\cos^3{\theta} = \frac{1}{4}(\cos{3\theta}+3\cos{\theta}) $$
Putting these into the equation gives
$$ 8\cos^{3}{\theta}-4\cos^2{\theta}-4\cos{\theta}+1 \\
= 2\cos{3\theta}+6\cos{\theta} -2\cos{2\theta}-2-4\cos{\theta}+1 \\
= -1+2(\cos{3\theta}-\cos{2\theta}+\cos{\theta}) = \frac{\cos{(7x/2)}}{\cos{(x/2)}}, $$
the last part of which comes from the formula
$$ \sum_{k=-n}^n (-1)^k \cos{kx} = (-1)^n \frac{\cos{(n+1/2)x}}{\cos{(x/2)}}, $$
which can be proven by induction. It's then clear that this is zero if $x$ is a zero of $\cos{(7x/2)}$, but not $\cos{(x/2)}$, and the first one of these is $\theta=\pi/7$.
A: We can use theory of equations (and some disguised Galois theory).
Given equation can be rewritten as $(2x)^3-(2x)^2 -2( 2x)+1=0$. So we need to show $2\cos \theta $ is a root of $x^3-x^2-2x+1=0$. We will do this by showing there is a cubic equation with integer coefficients satisfied by $2\cos\theta$, and determine the other two roots, and from that reconstruct the equation.
First let us write $\pi/7 = 2\pi/14=\theta$. Let $\alpha=e^{2\pi i/14}$, a primitive 14th root of unity. 
Clearly $-\alpha$ is a  primitive $7$th root of unity.  The latter is any solution of $x^6+x^5+x^4+x^3+x^2+x+1=0$. This tells us that $\alpha$ is a root of $x^6-x^5+x^4-x^3+x^2-x+1=0$.
Now  $\alpha+\bar\alpha = 2\cos\theta=\alpha+\alpha^{13}$. We will compute the polynomials satisfied by this number.
The Galois conjugates (the other roots) of $\alpha+\alpha^{13}$ are $\alpha^3+\alpha^{11},\alpha^5+\alpha^9$ (pair numbers less than 14 and coprime to it such that they add up to 14, and sum the corresponding powers of $\alpha$.)
It is a cubic with roots $a,b,c$ where $a=\alpha+\alpha^{13}, b=\alpha^3+\alpha^{11}, c=\alpha^5+\alpha^9$. So we need to calculate $a+b+c, ab+bc+ca$ and $abc$. So it boils down to showing $a+b+c=1, ab+bc+ca=-1, 
abc=-1$. Now this is a routine verification using the fact that $\alpha^{14}=1$,
