<s>[INST] What are the roots of the equation $8x^3-6x+1=0$? Let the roots be, $a,b$ and $c$
We have $$a+b+c=0\tag1$$ $$ab +bc +ca=-\frac68\tag2$$ $$abc = -\frac18\tag3$$
 [/INST] Note that 
$$\cos 3\theta=4\cos^3\theta-3\cos\theta$$
Let $x=\cos\theta$. Then
\begin{align*}
8\cos^3\theta-6\cos\theta+1&=0\\
2\cos3\theta+1&=0\\
\cos3\theta&=\frac{-1}{2}\\
3\theta&=2n\pi\pm\frac{2\pi}{3}\\
\theta&=\frac{2n\pi}{3}\pm\frac{2\pi}{9}
\end{align*}
So $x=\cos\frac{2\pi}{9}\approx 0.7660$, $\cos\frac{4\pi}{9}\approx 0.1736$ or $\cos\frac{8\pi}{9}=-0.9397$.
By letting $x=\cos\theta$, we can only find roots with $|x|\le1$. But as we already have $3$ roots, we have found all the roots.
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