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?
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 $C=\sqrt[3]{\frac{7+7\sqrt{27}i}2}$.