<s>[INST] Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7},\cos\frac{5 \pi}{7}$ 
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7}\space \text{and}\space \cos\frac{5 \pi}{7}$. 

How to even solve it ? I have no idea. 
Because cubic have no formula like quadratic equations to obtain the roots and the expression like $\cos\frac{\pi}{7}$ is something that I don't think I can evaluate with my limited knowledge (I'm in highschool) of only $\cos 2x$ or $\cos 3x$ type. Since it has denominator of 7, I thought maybe the seventh root of unity might help me, but since only some of its multiples are used, I don't think so it will be of much use. 
How do I solve this problem. Please help.
 [/INST] $$\cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}=\frac{8\sin\frac{\pi}{7}\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}{8\sin\frac{\pi}{7}}=\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}=-\frac{1}{8}.$$
The rest is a similar and use the Viete's theorem.
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