Prove $2\cos\frac{2\pi}{7}$ is root of $x^3 + x^2 - 2x -1$ To prove that a regular septagon cannot be constructed by a straightedge and compass, it suffices to prove that $2\cos(\frac{2\pi}{7})$ is not constructible. 

Several other answers to this question follow with the fact that $2\cos(\frac{2\pi}{7})$ is a root of 

$$x^3 + x^2 - 2x -1$$

How do I derive that? 

I know that since the cubic polynomial is irreducible in $\mathbb{Q}$, it has no real roots, implying that $cos(\frac{2\pi}{7})$ is not constructible.
Let $a = \frac{2\pi}7$. Then, $3a + 4a = 2\pi$ and 

$$\cos4a = \cos3a$$

Use the identities $\cos 2t = 2\cos^2t -1$ and $\cos3t = 4\cos^3 t-3\cos t$ to expand the two sides,

$$2(2\cos^2 a -1)^2 -1 = 4\cos^3a -3\cos a $$

Rearrage,

$$8\cos^4a -4\cos^3a -8\cos^2a +3\cos a+1=0$$

which factorizes as

$$(\cos a -1) (8\cos^3a + 4\cos^2a-4\cos a-1)=0$$

Since $\cos a \ne 1$, the equation above shows that $2\cos \frac{2\pi}7 $ satisfies 

$$x^3+x^2 -2x-1=0$$
Because $$\left(2\cos\frac{2\pi}{7}\right)^3+\left(2\cos\frac{2\pi}{7}\right)^2-2\left(2\cos\frac{2\pi}{7}\right)-1=$$
$$=8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=$$
$$=8\cos^3\frac{2\pi}{7}-6\cos\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-2+2\cos\frac{2\pi}{7}+1=$$
$$=2\cos\frac{6\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{2\pi}{7}+1=$$
$$=\frac{2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}}+1=$$
$$=\frac{\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}}{\sin\frac{\pi}{7}}+1=0.$$