Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the  whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think the symbols might be different but can be the same. If it is correct, please help me to solve this; if the equation is wrong, then please modify it and solve it.

$$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle\cos4x=-\cos3x=\cos(\pi+3x)\implies4x=2n\pi\pm(\pi+3x)$ where $n$ is any integer
'+' $\displaystyle\implies x=(2n+1)\pi\equiv\pi\pmod{2\pi}\implies\cos x=-1$
'-' $\displaystyle\implies x=\frac{(2n-1)\pi}7$
Now $\displaystyle\cos4x=-\cos3x$
Using Multiple angle formula,
$\iff8c^4-8c^2+1=-(4c^3-3c)\iff8c^4+4c^3-8c^2-3c+1=0\ \ \  \ (1)$ where $c=\cos x$
Clearly, the roots of $(1)$ are $\cos x,$ where $\displaystyle x=\pi, \frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$
So, the equation whose roots are $\displaystyle\cos\frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$ is  $$\frac{8c^4+4c^3-8c^2-3c+1}{c+1}=0$$
Here $n=2$
