Factoring cubic polynomial over R $z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4\cos^2(\frac{\pi}{14})$, $4\cos^2(\frac{3\pi}{14})$,$4\cos^2(\frac{9\pi}{14})$, but I don't know how to get them.
Thanks for help.

The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 \cos \frac{2 \pi}{7} \; , \; \; 2 \cos \frac{4 \pi}{7} \; , \; \;2 \cos \frac{8 \pi}{7} \; . \; \; $$
This is pretty easy if we take $\omega $ a 7th root of unity, then take $x = \omega + \frac{1}{\omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE

