Number of solutions of $\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) =0$ 
Solve in the interval $[0,2\pi]$ : $$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right)=0 $$

I tried expanding the L.H.S by applying the formula of $\cos(A+B)$ but it results in a quintic polynomial in terms of $\cos x $. Wolfram alpha has simplified the left hand side all the way to $\frac{15}{16}\cos3x$ but I'm unable to think of any method to reach there.

Use $\cos (A+B)$ formula to show for a natural number $n$,
$$\cos nx + \cos \left(nx+\frac{2n\pi}{3}\right) + \cos \left(nx+\frac{4n\pi}{3}\right) = \begin{cases} 0 , \quad \text{when n is coprime to 3} \\ 3\cos nx , \quad \text{when n is a multiple of 3} \end{cases}$$
Combining this with the identity from other answer,
$$\cos^5 \theta = \frac{1}{16}\cos 5\theta + \frac{5}{16}\cos  3\theta + \frac{10}{16}\cos  \theta$$
it is easy to see that first and third terms, $n=1,5$, should sum to zero, while middle term remains to give
$$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) = \frac{15}{16}\cos  3x$$
