Rational Polynomial of Degree $3$ satisfying $2\cos{(2\pi/7)}$ Let $\eta = \zeta_{7}+\bar{\zeta_{7}}$, for $\zeta_{7}=\exp{(2i\pi/7)}$. Find a polynomial of degree 3 with rational coefficients that $\eta$ satisfies.  

I'm not so sure on how to begin by this question. I saw somewhere that a way to do this would be looking at the automorphisms of $\mathbb{Q}(\zeta_{7})$, but I don't know how to apply this hint to this situation. I as well know that $\eta=2\cos{(2\pi/7)}$, not sure if this helps though.
What would be a way to get started for this problem? I'm lost.
Thanks for the help.
 A: The number $e^{\pi i/7}$ (and its reciprocal) are roots of $x^7+1=0$, and therefore of
$$x^6-x^5+x^4-x^3+x^2-x+1=0,$$
or equivalently of 
$$(x^3+x^{-3})-(x^2+x^{-2})+(x+x^{-1})-1=0\tag{1}$$
(we divided through by $x^3$ and rearranged).
Let $t=x+x^{-1}$. Then $x^3+x^{-3}=t^3-3t$, and $x^2+x^{-2}=t^2-2$. Substitute in (1) and simplify, and we obtain our cubic equation in $t$. 
Since $x=e^{\pm\pi i/7}$ are roots of (1), it follows that $t=e^{\pi i/7}+e^{-\pi i/7}$, that is, $2\cos(\pi/7)$, is a root of our cubic.
A: Pretty much a brute force approach:
$$\begin{align}\eta^2 &= \zeta_7^2 + 2 + \overline\zeta_7^2\\
\eta^3 &= \zeta_7^3+3(\zeta_7+\overline\zeta_7) +\overline\zeta_7^3\end{align}$$
Now remember that $\overline\zeta_7=\zeta_7^6,\overline{\zeta_7}^2=\zeta_7^5,\overline\zeta_7^3=\zeta_7^{4}$.
So: $$\eta^2+\eta^3 = 1+(\sum_{i=0}^6 \zeta_7^i) + 2\eta=1+2\eta$$
Alternatively, $\cos 2\pi/7$ is a double root of $T_7(x)-1$, where $T_7(x)$ is the seventh Chebyshev polynomial of the first kind.  Then $\cos 2\pi/7$ is a root of the GCD of $T_7(x)-1$ and $T_7'(x)$, which is $8x^3+4x^2-4x-1$. Then $2\cos2\pi/7$ is a root of $x^3+x^2-2x-1$.
Finally, you could just realize the minimal polynomial must be:
$$(x-\zeta-\zeta^6)(x-\zeta^2-\zeta^5)(x-\zeta^3-\zeta^4)$$
And that the coefficients must be integers since they are fixed by automorphisms of the original field. You can work them out - it isn't hard.
