Problem. Prove that there exists a continuous function $f: \mathbb{R}\rightarrow \mathbb{R}$ so that $$4 f(\!x\!)\!f(\!x+ \frac{\pi}{3}\!)\!f(\!x+ \frac{2 \pi}{3}\!)\!-\!f^{3}(\!x\!)\!-\!f^{3}(\!x+ \frac{\pi}{3}\!)\!-\!f^{3}(\!x+ \frac{2 \pi}{3}\!)\!=\!\cos^{2} x\!+\!\cos^{2} (\!x+ \frac{\pi}{3}\!)\!+\!\cos^{2} (\!x+ \frac{2 \pi}{3}\!)$$ I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
Solution. We have $$\cos^{2} x+ \cos^{2} (\!x+ \frac{\pi}{3}\!)+ \cos^{2} (\!x+ \frac{2 \pi}{3}\!)= \cos^{2} x+ (\!\frac{1}{2}\cos x- \frac{\sqrt{3}}{2}\sin x\!)^{2}+ (\!\frac{1}{2}\cos x+ \frac{\sqrt{3}}{2}\sin x\!)^{2}$$ $$= \cos^{2} x+ \frac{1}{2}\cos^{2} x+ \frac{3}{2}\sin^{2} x= \frac{3}{2}\,\forall\,x$$ If $f(x)= f(\!x+ w\dfrac{\pi}{3}\!)\,(\!w\in \mathbb{Z}\!)$ then the left side is $[f(x)]^{3}$. We can choose $f(x)= \sqrt[3]{\dfrac{3}{2}}$ be a solution .
q.e.d