If $\zeta= e^{2\pi i/7}$ then $2\cos(2\pi/7) = (\zeta+ \zeta^6)$. Find a cubic polynomial satisfied by $\zeta + \zeta^6$. My approach was to set $\zeta + \zeta^6 = a_1x + a_2x^2 + a_3x^3$ and then solve for the constants.
I know there is a better way to solve this using the cyclotomic polynomials, but I don't understand how to do that. If anyone is willing to explain I would appreciate it. 
 A: You know that $\zeta^7=1$, so $\zeta$ is a root of $X^6+X^5+X^4+X^3+X^2+1$. Moreover $\zeta^6=\zeta^{-1}$. Consider
$$
\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0
$$
and divide by $\zeta^3$:
$$
\zeta^3+\frac{1}{\zeta^3}+\zeta^2+\frac{1}{\zeta^2}+\zeta+\frac{1}{\zeta}+1=0
$$
Observe that
$$
x^3+y^3=(x+y)^3-3xy(x+y)
$$
and
$$
x^2+y^2=(x+y)^2-2xy
$$
Use with $x=\zeta$ and $y=\zeta^{-1}$.

 $\left(\zeta+\dfrac{1}{\zeta}\right)^3-3\left(\zeta+\dfrac{1}{\zeta}\right)+\left(\zeta+\dfrac{1}{\zeta}\right)^2-2+\left(\zeta+\dfrac{1}{\zeta}\right)+1=0$

A: I do it by a method that’s the same as that of @egreg, but arranged differently.
First I write out
$$
0=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}\,.
$$
Then I write $\xi=\zeta+\zeta^{-1}$ and write out
$$
\xi^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}\,,
$$
and subtract zero (from the first display) from it, to get the equation
$$
\xi^3=-\zeta^2+2\zeta-1+2\zeta^{-1}-\zeta^{-2}\,,
$$
and this is an obvious linear combination of $\xi^2$, $\xi$, and $1$.
