We know that $cos^n(x)$ is a linear combination of $1,cos(x),cos(2x),\dots,cos(nx)$. The expression on the left is invariant under $x \to x+ 2\pi/3$ so that only $1, cos(3x), cos(6x)$ appear. The expression changes sign under $x \to x+\pi$ so that only $cos(3x)$ has a nonzero coefficient. So the problem reduces to evaluation of the sum at some particular $x$, such as $x=0$, where the cosines are $1, -1/2, -1/2$ and the computation is trivial.

fleeting_guest, thank for your good solution I found another good solution for very ugly identity: solution: Suppove that $A_n(x)=\cos^nx+\cos^n(x+\frac {2\pi}{3})+\cos^n(x+\frac {4\pi}{3})$ . By useing the equation $4\cos^3x=\cos {3x}+\cos x$ so that $A_n(x)=\frac {\cos {3x}}{4}A_{n-3}(x)+\frac 34 A_{n-2}(x)$ then we will have $A_0(x)=3 , A_1(x)=0 , A_2(x)=\frac 32 , A_3(x)=\frac 34 \cos {3x} , A_4(x)=\frac 98 , A_5(x)=\frac {15}{16}\cos {3x}, A_7(x)=\frac {63}{64}\cos {3x}$