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?

Use the product-to-sum identities on the RHS a couple of times:
\begin{align}
  4\sin\frac{B-C}{2}\sin\frac{C-A}{2}\sin\frac{A-B}{2}
      &= 4\cdot\frac{1}{2}\left(\cos\frac{B+A-2C}{2} - \cos\frac{B-A}{2}\right)\sin\frac{A-B}{2} \\
      &= 2\left(\cos\frac{B+A-2C}{2}\sin\frac{A-B}{2} - \cos\frac{B-A}{2}\sin\frac{A-B}{2}\right) \\
      &= \sin(A-C) - \sin(B-C) - \sin 0 + \sin(B-A) \\
      &= \sin(A-C) + \sin(C-B) + \sin(B-A).
\end{align}
Then, as you observed in the original post, since $A+B+C=\pi$, this is equal to
$$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B).$$
