Croatia competition problems

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
$$\cos \frac{x}{3} + \cos \frac{x+2\pi}{3} + \cos \frac{x+4\pi}{3} = 2 \cos \frac{2\pi}{3} \cos \frac{x+2\pi}{3} + \cos \frac{x+2\pi}{3} \\ = \left( 2 \cos \frac{2\pi}{3} + 1 \right) \cos \frac{x+2\pi}{3} = 0.$$
Finally, we conclude
$$\cos^3 \frac{x}{3} + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x.$$