Trigonometric relation between sides and angles of a triangle $$a \cdot \sin (B-C) +b \cdot \sin(C-A) +c \cdot \sin(A-B) =0$$
where $a, b, c$ are the sides of a triangle and $A, B, C$ are the angles of a triangle.
No idea how to solve this problem.
 A: $$
a \sin(B-C) + b \sin(C-A) + c \sin(A-B)\\
= a ( \sin B \cos C - \sin C \cos B) + b(\sin C \cos A - \sin A \cos C) + c (\sin A \cos B - \sin B \cos A)\\
= a \sin B \cos C - a \sin C \cos B + b\sin C \cos A - b \sin A \cos C + c \sin A \cos B - c \sin B \cos A\\
= a \sin B \cos C - b \sin A \cos C + b \sin C \cos A  - c \sin B \cos A + c \sin A \cos B - a \sin C \cos B 
$$
The sine rule says that
$$
\frac{\sin A }{a} = \frac{\sin B }{b} = \frac{\sin C}{c}
$$
So we have:
$$
b\sin A  = a\sin B \\
c\sin B = b\sin C\\
a\sin C = c \sin A
$$
Substituting this into the final expression from above gives:
$$
a \sin(B-C) + b \sin(C-A) + c \sin(A-B)\\
= a \sin B \cos C - b \sin A \cos C + b \sin C \cos A  - c \sin B \cos A + c \sin A \cos B - a \sin C \cos B \\
= a \sin B \cos C - a \sin B \cos C + b\sin C \cos A - b \sin C \cos A + c\sin A \cos B - c \sin A \cos B\\
=0
$$
A: Using Law of Sines, $\displaystyle a=2R\sin A$
Again, $\displaystyle\sin A=\sin[\pi-(B+C)]=\sin(B+C)$
$$\implies a\sin(B-C)=2R\sin A\sin(B-C)=2R\sin(B+C)\sin(B-C)$$
Now use Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$
