Factor $z^7-1$ into linear and quadratic factors and prove that $$ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$$
I have been able to prove it using the value of $\cos(\pi/7)$. Given here http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
Let $z=e^{\frac{i\pi}{7}}$. Then $\cos (\frac{\pi}{7})=(z+z^{-1})/2$, $\cos (\frac{2\pi}{7})=(z^2+z^{-2})/2$, $\cos (\frac{3\pi}{7})=(z^3+z^{-3})/2$. This should get you started.
Factor $x^7-1$ in $\Bbb C$ and obtain its factorization in $\Bbb R$ by pairing off conjugate roots.
Let $$\alpha_1 = \cos(2\pi/7), \alpha_2 = \cos(4 \pi/7), \alpha_3 = \cos(6 \pi/7)$$
Then $$ z^7-1 = (z-1) (z^2- 2 \alpha_1 z + 1)(z^2- 2 \alpha_2 z + 1)(z^2- 2 \alpha_3 z + 1)$$
Differentiate both sides and set $z=1$ to get your answer.
$$z^7=1=e^{2n\pi i}$$ where $n$ is any integer
$$\implies z=e^{\frac{2n\pi i}7}$$ where $n=0,\pm1,\pm2,\pm3$
So, the roots of $$\frac{z^7-1}{z-1}=0\iff z^6+z^5+\cdots+z+1=0\quad(1)$$ are $e^{\frac{2n\pi i}7}$ where $n=\pm1,\pm2,\pm3$
As $z\ne0,$ divide either sides by $z^3$ to get $$z^3+\frac1{z^3}+z^2+\frac1{z^2}+z+\frac1z+1=0\quad(2)$$
Now using Euler Formula, $\displaystyle z+\frac1z=e^{\frac{2n\pi i}7}+e^{-\frac{2n\pi i}7}=2\cos\frac{2n\pi }7$ where $n=1,2,3$
Again, $$\displaystyle z^2+\frac1{z^2}=\left(z+\frac1z\right)^2-2\text{ and }z^3+\frac1{z^3}=\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)$$
Put the values of $\displaystyle z^2+\frac1{z^2},z^3+\frac1{z^3}$ in $(2)$ to form a Cubic Equation whose roots are $\displaystyle2\cos\frac{2n\pi }7$ where $n=1,2,3$
Now apply Vieta's formula
