prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$ If $A + B + C = \pi$,
then prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$
My try:

$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = \sin(A - C) + \sin(B - A) + \sin(C -B)$
$\sin(A - C) + \sin(B - A) + \sin(C -B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \cos(A - \frac{C + B}{2}) + \sin(C - B)$
$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \sin(\frac{A}{2}) + \sin(C - B)$

I am confused what steps should follow?

Another way to do it, is to use exponential notation.
(1) First, as $A+B+C=\pi$ you can simplify your LHS to
$$\sin(B+2C)+\sin(C+2A)+\sin(A+2B)\\=\sin(\pi-A+C)+\sin(\pi-B+A)+\sin(\pi-C+B)\\
=-\sin(C-A)-\sin(A-B)-\sin(B-C)$$
Now our problem only depends on the distances of the three values, so we can reduce its dimension by substituting
$$A-B=x,\quad B-C=y, \quad\Rightarrow C-A=-x-y$$ and obtain an identity of only two variables $x,y\in\mathbb{R}$ which is
$$\sin(x+y)-\sin(x)-\sin(y)=-4\sin\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\sin\left(\frac{x+y}{2}\right).$$
(2) Now we can proof this equation using exponential identity $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and simple exponential algebra. The RHS is
$$=\underbrace{\frac{-4}{(2i)^3}}_{\displaystyle{\frac{1}{2i}}}\underbrace{\underbrace{\left(e^{ix/2}-e^{-ix/2}\right)
\left(e^{iy/2}-e^{-iy/2}\right)}_{\displaystyle{\left(e^{i(x+y)/2}-e^{i(x-y)/2}-e^{i(y-x)/2}+e^{-i(x+y)/2}\right)}}
\left(e^{i(x+y)/2}-e^{-i(x+y)/2}\right)}_{\displaystyle{\left(e^{i(x+y)}-e^0-e^{ix}+e^{-iy}-e^{iy}+e^{-ix}+e^0-e^{-i(x+y)}\right)}}$$
which can obviously grouped back to form
$$=\sin(x+y)-\sin(x)-\sin(y).$$
