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.

• You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ? – Yves Daoust Nov 24 '18 at 19:44
• It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha. – mrtaurho Nov 24 '18 at 19:45
• Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post. – Dovla Nov 24 '18 at 19:46
• @YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $\cos^2 \dfrac {n \pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $\cos^2 \dfrac {\pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how. – Ovi Nov 24 '18 at 20:20
• @ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano. – Yves Daoust Nov 24 '18 at 20:38

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.

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