If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$, 
If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that
  $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$


My solution:
From the given, 
$$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$
Now, 
L.H.S$$=\cos3A+\cos3B+\cos3C$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$
$$=12\cos A \cos B \cos C.$$
My solution ends here. How should I complete?
NOTE: I am not allowed to use complex numbers at my level. So, please help me solve this problem without using complex numbers.

$\cos (A+B+C)=\cos (A+B) \cos C- \sin (A+B) \sin C \\ = (\cos A \cos B -\sin A \sin B) \cos C -( \sin A \cos B +\cos A \sin B) \sin C \\ =\cos A \cos B \cos C- \sum_{cyc} \sin A \sin B \cos C $
$( \sin A+ \sin B)^2= (-\sin C)^2 , (\cos A +\cos B )^2=(-\cos C)^2  \\ \implies 2+ 2\sin A \sin B +2 \cos A \cos B =1 \\ \implies  \sin A \sin B = -\cos A \cos B -\dfrac{1}{2} \\ \implies \sum_{cyc} \sin A \sin B \cos C =\sum_{cyc} (-\cos A \cos B -\dfrac{1}{2})\cos C  \\ =-3 \cos A \cos B \cos C -\dfrac{1}{2} (\cos A + \cos B + \cos C)=-3 \cos A \cos B \cos C$
