What are the roots of the equation $8x^3-6x+1=0$?

Question

Let the roots be, $a,b$ and $c$

We have $$a+b+c=0\tag1$$ $$ab +bc +ca=-\frac68\tag2$$ $$abc = -\frac18\tag3$$

Answer 1

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.

Answer 2

Not solvable by hand? I need hands only to write down the results: $x_1=\cos\frac{2\pi}{9},x_2=\cos\frac{7\pi}{9},x_3=\cos\frac{13\pi}{9}$. Simply observe $\cos3\phi=4\cos^3\phi-3\cos\phi$.