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

$\begin{array}{l}\cos A+\cos B+\cos C=0=\sin A+\sin B+\sin C\\ x=cosA+i\sin A,y=\cos B+i\sin B,z=\cos C+i\sin C\\ x+y+z=0\\ x+y=-z\\ {\left(x+y\right)}^3=-z^3\\ x^3+y^3+3xy\left(x+y\right)=-z^3\\ x^3+y^3+3xy\left(-z\right)=-z^3\\ x^3+y^3+z^3=3xyz\\ {\left(cisA\right)}^3+{\left(cisB\right)}^3+{\left(cisc\right)}^3=3cisAcisBcisC\\ cis3A+cis3B+cis3C=3cis(A+B+C)\\ comparing real parts on both sides\\ \cos 3A+\cos 3B+\cos 3C=3\cos \left(A+B+C\right)\\ \end{array}$