If $A+B+C=\pi$, prove: $\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$ 
If $A+B+C=\pi$, prove that:
  $$\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$$

My Attempt:
Here, $A+B+C=\pi$
Now,
$$\begin{align}
LHS &=\cos(B+2C)+\cos(C+2A)+\cos(A+2B) \\
&=\cos(B+C+C)+\cos(C+A+A)+\cos(A+B+B) \\
&=\cos(\pi-(A-C))+\cos(\pi-(B-A))+\cos(\pi-(C-B)) \\
&=-\cos(A-C)-\cos(B-A)-\cos(C-B)
\end{align}$$
Please help to continue from here.

Let $C-A=2x,B-C=2y,A-B=2z\implies2(x+y+z)=0$
$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$
Now as $\cos(x+y)=\cos(-z)=\cos z,$
$$F=2\cos z\cos(x-y)+2\cos z\cdot\cos(x+y)-1$$
$$=2\cos z\{\cos(x+y)+\cos(x-y)\}-1=2\cos z\{2\cos x\cos y\}-1=?$$
