Show that $2\cos \frac{2\pi}{7}$ satisfies $x^3+x^2-2x+1$

Question

Show that $2\cos \frac{2\pi}{7}$ satisfies $x^3+x^2-2x+1$

I need to use Galois Theory. I know about the identity $\cos 3\theta = 4\cos^3 -\cos \theta$, but I can't fit things into the polynomial.

Answer 1

Consider $\frac{x^7-1}{x-1}=\Phi_7(x)=1+x+x^2+x^3+...+x^6$

It has his zeros in $\exp(\frac{2\pi i k}{7})$ with $k \in [0,1]$

If $\alpha$ is a root of $\Phi_7$ then $$\left(\alpha^3 +\frac{1}{\alpha^3}\right)+\left(\alpha^2 +\frac{1}{\alpha^2}\right)+\left(\alpha+\frac{1}{\alpha^1}\right)+1=0$$ $$\left(\alpha^2 +\frac{1}{\alpha^2}\right)=\left(\alpha +\frac{1}{\alpha}\right)^2-2$$ $$\left(\alpha^3 +\frac{1}{\alpha^3}\right)-\left(\alpha +\frac{1}{\alpha}\right)\left(\alpha^2-1 +\frac{1}{\alpha^2}\right)=\left(\alpha^3 +\frac{1}{\alpha^3}\right)-3\left(\alpha +\frac{1}{\alpha}\right)$$

Consider $x=\left(\alpha +\frac{1}{\alpha}\right)$ and you obtain: $$x^3-3x+x-2+x^2+1=0$$

At the end you obtain $$x^3+x^2-2x-1=0$$

When $\alpha=\exp\left(\frac{2\pi i}{7}\right)$ you get $x=2\cos\left(\frac{2\pi}{7}\right)$

Answer 2

You could use $\cos7\theta=T_7(\cos\theta)$ where $T_7$ is the seventh Chebyshev polynomial of the first kind (q.v.).

Or just write $$2\cos\frac{2\pi}{7}=\newcommand{\ze}{\zeta}\ze+\ze^{-1}$$ where $\zeta=\exp(2\pi i/7)$ and compute $$x^3+x^2-2x-1=(\ze+\ze^{-1})^3+(\ze+\ze^{-1})^2-2(\ze+\ze^{-1})-1 =\ze^3+\ze^2+\ze+1+\ze^{-1}+\ze^{-2}+\ze^{-3}=\frac{\ze^4-\ze^{-3}}{\ze-1} =0$$ as $\ze^7=1$.

Answer 3

If $7x=2m\pi$ where $m$ is any integer

$4x=2m\pi-3x$

$\cos4x=\cos3x$

$2(2c^2-1)^2-1=4c^3-3c$ where $c=\cos x$

$\iff8c^4-4c^3-8c^2+1=0$

Now if $7\mid m,c=1$

So, if $7\nmid m$ $$\dfrac{8c^4-4c^3-8c^2+1}{c-1}=0$$