Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2\cos(\frac {2\pi}{9})-1]$$.
What should I do further?

$2(4 \cos^3 \frac π9 - 3 \cos \frac π9)$
As $\cos 3\theta = 4 \cos^3 \theta - 3\cos \theta$
= $2(\cos (3 × \frac π9))$
=$2 × \cos(\frac π3)$
=$2 × \frac 12 = 1$
