<s>[INST] 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.
 [/INST] For the RHS, you can use a known formula for the product of three cosines (see eg. here): 
$\cos x \; \cos y \; \cos z = \frac{1}{4} \Big(\cos (x+y-z) + \cos (y+z-x) + \cos (z+x-y) + \cos (x+y+z)\Big)$
Considering your RHS, you set $x = \frac{B-C}{2}$,  $y = \frac{C-A}{2}$,  $z = \frac{A-B}{2}$,    and plug it into the formula. Then, for example, 
$x+y-z = B-A$, likewise the other arguments.
This gives immediately the form you arrived at,
$-\cos(A-C)-\cos(B-A)-\cos(C-B)$.
</s>