How would you find the roots of $x^3-3x-1 = 0$ I'm not too sure how to tackle this problem. Supposedly, the roots of the equation are $2\cos\left(\frac {\pi}{9}\right),-2\cos\left(\frac {2\pi}{9}\right)$ and $-2\cos\left(\frac {4\pi}{9}\right)$
How do I start? The Cosines seem especially scary...

HINT:
Let $x=b\cos A$
$$\implies b^3\cos^3A-3(b\cos A)=1$$
As $\cos3A=4\cos^3A-3\cos A,$
$$\dfrac43=\dfrac{b^3}{3b}\implies b^2=4\text{  as }b\ne0$$
Let $b=2$
Consequently, $$2\cos3A=1\iff\cos3A=\dfrac12=\cos\dfrac\pi3$$
$$3A=2n\pi\pm\dfrac\pi3=\dfrac\pi3(6n\pm1)$$ where $n$ is any integer
$$A=\dfrac\pi9(6n\pm1)$$ where $n\equiv0,\pm1\pmod3$
What if $b=-2 :)$
