<s>[INST] Equation $(8\cos^3x+1)^3=162\cos x-27$ 
Solve equation $$(8\cos^3x+1)^3=162\cos x-27$$

I saw this equation before 5 month, and I couldn't solve it. This isn't homework, etc. (I don't do stuff like this anymore). I am just curious. 
 [/INST] This approach is based off knowing that the answers are $ \frac {2 \pi}{9} , \frac {4 \pi }{9} , \frac { 8 \pi }{9} $. Another equation whose solutions are exactly those values, is $ 2 \cos 3 x  + 1 = 0$, so we'd look to factorize that out.
We know that $ 2 \cos 3x + 1 = 8\cos^3 - 6 \cos x + 1$, so  $ (8 \cos^3 x + 1) = (2 \cos 3x + 1 + 6 \cos x)$. Henceforth, let $w = 2 \cos 3x + 1$.
We have $ ( w + 6 \cos x) ^3 = 162 \cos x - 27$, or that $w^3 + 3 w^2 6\cos x  + 3 w (6 \cos x)^2 + 216 \cos^3 x = 162 \cos x - 27$.
Continue to focus on factorizing out $w$, we get
$w^3 + 18 w^2 \cos x + 108 w \cos^2x = - 216 \cos^3 x + 162 \cos x - 27 = - 27 w$, 
hence
$w^3 + 18w^2 \cos x + 108 w \cos^2x + 27 w = 0$.
This factorizes into $0 = w ( w^2 + 18 w \cos x + 108 \cos^2 x + 27) = w [ (w + 9 \cos x)^2 + 27 \cos^2 x + 27 ]. $
It is clear that the other term is strictly positive, so has no solution for $x$ (regardless of what $w$ is), so we must have $w = 0$. Hence, the solutions are $x = \frac {2\pi}{9}, \frac { 4\pi}{9}, \frac {8\pi}{9}$.
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