$\mathbf{f(x)=\cos^3 2x= \frac{1}{2}a_0+\sum^{\infty}_{n=1}(a_n \cos nx + b_n \sin nx)}$
Since $\cos^3 2x$ is even $\rightarrow b_n=0$
$a_0=\frac{1}{2\pi}\int^{\pi}_{-\pi} \cos^3 2x$ $dx$
$a_0=\frac{1}{\pi}\int^{\pi}_{0} \frac{1}{4}(\cos 6x+3 \cos 2x)$ $dx$
$a_0=\frac{1}{4\pi}[\frac{\sin 6x}{6}+\frac{3\sin 2x}{2} ]^{\pi}_{0}=0$
$a_n=\frac{1}{4\pi}\int^{\infty}_{-\infty}(\cos 6x \cos nx)$ $dx$ $\frac{3}{4\pi}\int^{\infty}_{-\infty}(\cos 2x \cos nx)= \frac{1}{4\pi}(\pi)+\frac{3}{4\pi}(\pi)=1$
I get $\cos^3 2x=\sum^{\infty}_{n=1} \cos nx$
The answer is
$\cos^3 2x=\frac{1}{4}(3\cos 2x+ \cos 6x)$