Show that $Q(cos(40°))$ a normal extension of $Q$

Question

I've calculated the minimal polynomial of $cos(40°)$ is $p(x) = 8x^3 - 6x + 1$. Since it has a degree of 3, the extension is finite. Also $(p', p) = 1$ so it has no multiple roots. I've tried writing $ak^2 + bk +c$ into $p$ but i got back $k$ or some irrational $=c$. Also tried showing that $Q(k, l, m) = Q(k)$ with no luck. What am I missing? ($k, l, m$ are the roots of $p$, $k = cos(40°)$

Answer 1

The other roots of your cubic are $\cos80^\circ$ and $\cos160^\circ$. For $\gamma=\cos40^\circ$, $2\gamma^2-1=\cos80^\circ\in\Bbb Q(\gamma)$. Likewise $\cos160^\circ\in\Bbb Q(\gamma)$. As $p$ splits in $\Bbb Q(\gamma)$, it's a normal extension of $\Bbb Q$.