If $A+B+C=\pi$ and $\cos A+\cos B+\cos C=0=\sin A+\sin B+\sin C$, then $\cos 3A+\cos 3B+\cos 3C=$

$\begin{array}{l}a=cisA,b=cisB,c=cisC\\ a+b+c=cisA+cisB+cisC\\ =(\cos A+\cos B+\cos C)+i(\sin A+\sin B+\sin C)\\ a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\\ (\mathrm{cis}A{)}^3+(\mathrm{cis}B{)}^3+(\mathrm{cis}C{)}^3=3\text{ cisAcisBcisC }\\ cis3A+cis3B+cis3C=3cis(A+B+C)\\ (\cos 3A+\cos 3B+\cos 3C)+i(\sin 3A+\sin 3B+\sin 3C)\\ =3\cos (A+B+C)+i3\sin (A+B+C)\end{array}$

Equating real and imaginary part

$\begin{array}{l}\cos 3A+\cos 3B+\cos 3C=3\cos (A+B+C)\\ \cos 3A+\cos 3B+\cos 3C=3\cos \pi \\ \cos 3A+\cos 3B+\cos 3C=-3\end{array}$