I was asked to re-express the following number:
$$ z=e^{2\pi i/7}=\mathrm{cis}\left(2\pi/7\right) $$ using only numbers from ℚ, arithmetic operations and the cubic root operation.
I tried to follow the suggested solution here which I think it needs a minor modification, but unfortunately I'm stuck.
Any help, or a direction is welcomed.
Since $z^7=1$, $z\neq1$, and $z^7-1=(z-1)(z^6+z^5+z^4+z^3+z^2+1+1)$, $z$ is such that $z^6+z^5+z^4+z^3+z^2+1+1=0$. Let $w=z+\frac1z$. Then\begin{align*}z^6+z^5+z^4+z^3+z^2+1+1=0&\iff z^3+z^2+z+1+\frac1z+\frac1{z^2}+\frac1{z^3}=0\\&\iff w^3+w^2-2w-1=0.\end{align*}You can solve this equation using Cardano's formula and then solve the quadratic equation $z+\frac1z=w$.