Make $ x=2cos\alpha$.So, ${ 8cos^{3} \alpha}-6cos\alpha-1 = 0$. Therefore, $ cos3\alpha=\frac{1}{2}$.From this, we get that: $ x_{1}=2cos140 ; x_{2}=2cos80 ; x_{3}=2cos20$ So, we have to prove that: $ 2(cos20-cos80)(cos20+cos80)=cos20+cos40$. It's equivalent to: $ 2sen100sen60=2cos30cos10$ ; which is clearly true because $ sen100=sen80=cos10$. It finishes the proof. PD: The angles are in degrees.

You are asuming that $ -2 \leq x \leq 2$

Yes, but how I found three solutions for a third degree equation, I have found all of them. Thanks for your reply.

Let $f(x)=x^3 - 3x -1$. Examining the values of f(1), f(0), f(-1) and f(-2), we can see that $x_1 < x_2 < 0 < x_3$. Note that if $z$ is a root of $f(x)=0$, then $1/z$ and $(-1-z)$ are both roots of $g(x) = x^3+3x^2-1=0$. But $1/x_2 <1/x_1 < 0 < 1/x_3$ and $-1-x_3 < -1-x_2 < -1-x_1$, so we must have $x_3+1/x_2= -1 = x_1+1/x_3$. Since $f(z)=0$ implies $z^2 = 3 + 1/z$, we can substitute and rearrange to get the required equality.