How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\ 
  (B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9}  \\ 
  (C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\ 
  (D)-2\cos\frac{2\pi}{11},2\cos\frac{8\pi}{11},2\cos\frac{14\pi}{11}$$
I tried to eliminate options using the sum and product of roots but I can't figure out if $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$$
or 
$$\cos\frac{2\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{14\pi}{11}=0$$

Put $x=2t$
Hence $$8t^3-6t+1=0$$ 
Now put $t=\cos \theta$
Hence $$2\cos (3\theta)=-1$$
And the rest is simple trigonometric equation
