Croatia_2018

Using the identity $$\cos t=4{\cos }^3\frac{t}{3}-3\cos \frac{t}{3}$$ for $t=x$, $t=x+2\pi$ and $t=x+4\pi$, summing the relations and using the periodicity of cosine, we obtain $$3\cos x=4\left({\cos }^3\frac{x}{3}+{\cos }^3\frac{x+2\pi }{3}+{\cos }^3\frac{x+4\pi }{3}\right)-3\left(\cos \frac{x}{3}+\cos \frac{x+2\pi }{3}+\cos \frac{x+4\pi }{3}\right)$$ Transforming the expression in the second parenthesis yields $$\begin{gathered} \cos \frac{x}{3}+\cos \frac{x+2\pi }{3}+\cos \frac{x+4\pi }{3}=2\cos \frac{2\pi }{3}\cos \frac{x+2\pi }{3}+\cos \frac{x+2\pi }{3} \\ =\left(2\cos \frac{2\pi }{3}+1\right)\cos \frac{x+2\pi }{3}=0. \end{gathered}$$ Finally, we conclude $${\cos }^3\frac{x}{3}+{\cos }^3\frac{x+2\pi }{3}+{\cos }^3\frac{x+4\pi }{3}=\frac{3}{4}\cos x.$$