How to solve the cubic $x^3-3x+1=0$?

Question

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 $$\cos\frac{2\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{14\pi}{11}=0$$

Answer 1

Hint: $(2\cos \theta)^3-3(2\cos\theta)+1=2\cos3\theta+1$

Answer 2

Let $x=a\cos t$

Comparing with $\cos3t$ formula $$\dfrac{a^3}4=\dfrac{3a}3, a=?$$

$$a\cos3t=-1$$

$$3t:360^\circ n\pm120^\circ$$

Answer 3

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

Answer 4

To follow on from your method, $\cos(a) + \cos(b) = 2 \cos({a+b\over2}) \cos({a-b\over2})$, so $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}=2 \cos\frac{5\pi}{9}\cos\frac{3\pi}9=\cos\frac{5\pi}{9}$$ but $\cos\frac{5\pi}{9} = -\cos\frac{14\pi}{9}$, so the sum $\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$.