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.
 A: 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

A: Let $z=aw+b$. The equation turns to
$$a^3w^3+(3ba^2-7a^2)w^2+(3ab^2-14ab+14a)w+b^3-7b^2+14b-7=0.$$
Now, cancel the quadratic coefficient. This is achieved by $3b=7$, and
$$a^3w^3-\frac73aw+\frac7{27}=0.$$
Next, ensure that the ratio of the cubic coefficient over the linear one is $-\dfrac43$. This is achieved by
$$a=\frac{\sqrt{28}}3,$$
and
$$\frac{28}9\frac{\sqrt{28}}3w^3-\frac73\frac{\sqrt{28}}3w+\frac7{27}=0$$ or
$$4w^3-3w=-\frac1{\sqrt{28}}.$$
Finally, setting
$$w=\cos\theta,$$
$$4\cos^3\theta-3\cos\theta=\cos 3\theta=-\frac1{\sqrt{28}}$$ gives you the roots.
