Solving the sextic equation with 14th root of unity I am solving the sextic equation $t^6-t^5+t^4-t^3+t^2-t+1=0$ satisfied by the 14th root of unity (a problem from Ian Stewart's book). I was able to get up to the point where you have the polynomial $u^3-u^2-2u+1=0$. But I got stuck and so I referred to the below question. 
Solving the sextic $t^6-t^5+t^4-t^3+t^2-t+1$
In it, the answer says that it is the minimal polynomial for $2\cos(2\pi/14)$, which I think is referring to the fact that the original polynomial satisfies the 14th root of unity. Can someone confirm this for me?
Further the answer continues saying that the $u$ can be written as a combination of the third root of unity and $C=\sqrt[3]{\frac{7+7\sqrt{27}i}2}$ but I am not sure where this comes from (unless this is another way to write the 14th root of unity). Can someone explain how they got there?
 A: Because the sextic polynomial is palindromic, finding roots can be reduced to finding roots of the corresponding cubic in $t+t^{-1}$. A few computations show that for $u=t+t^{-1}$ we have
\begin{eqnarray*}
u^3&=&(t+t^{-1})^3=t^3+3t+3t^{-1}+t^{-3}\\
u^2&=&(t+t^{-1})^2=t^2+2+t^{-2},
\end{eqnarray*}
which shows that
\begin{eqnarray*}
t^6-t^5+t^4-t^3+t^2-t+1
&=&t^3\Big((t+t^{-1})^3-(t+t^{-1})^2-2(t+t^{-1})+1\Big)\\
&=&t^3(u^3-u^2-2u+1).
\end{eqnarray*}
Note that for every positive integer $n$ you have
$$\zeta_n+\zeta_n^{-1}=2\cos(2\pi/n),$$
which immediately shows that $2\cos(2\pi/14)$ is a root of the cubic, because $\zeta_{14}$ is a root of the sextic. Reversing the argument above also shows that if this cubic is not the minimal polynomial of $2\cos(2\pi/14)$, then the sextic is not the minimal polynomial of $\zeta_{14}$.
Now Cardano's formula yields explicit expressions for the roots $\alpha_1$, $\alpha_2$ and $\alpha_3$ of the cubic
$$u^3-u^2-2u+1=0,$$
and then solving $t+t^{-1}=\alpha_k$, which simplify to quadratic polynomials in $t$, yields the roots of the sextic. I haven't done the computations myself, but I assume that plugging the coefficients of the cubic above into Cardano's formula yields
$$u=\frac13\left(1-\omega^kC-\frac7{\omega^kC}\right),\qquad k\in\{0,1,2\},$$
where $C=\sqrt[3]{\frac{7+7\sqrt{27}i}2}$.
