<s>[INST] 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.
 [/INST] we have $$sinA+sinB=-sinC$$ and $$cosA+cosB=-cosC$$ squaring and adding both we get
$$2+2cos(A-B)=1$$ that is
$$cos(A-B)=\frac{-1}{2}$$ $\implies$
$$A-B=\frac{2\pi}{3}$$ similarly
$$B-C=\frac{-4\pi}{3}$$ and
$$C-A=\frac{2\pi}{3}$$
Now $$cos3A=cos\left(3\left(B+\frac{2\pi}{3}\right)\right)=cos(3B+2\pi)=cos3B$$
similarly $$cos3C=cos3B$$
Hence
$$cos3A+cos3B+cos3C=3cos3A=3cos(A+A+A)=3cos\left(A+B+\frac{2\pi}{3}+C-\frac{2\pi}{3}\right)=3cos(A+B+C)$$
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