Prove $(b-c)\sin A+(c-a)\sin B+(a-b)\sin C=0$

Question

Prove the following equation, when you consider it as $BC=a$, $CA=b$, and $AB=c$ in a triangle　$ABC$.

$(b-c)\sin A+(c-a)\sin B+(a-b)\sin C=0$

Answer 1

Using this or this, $$\frac b{\sin B}=\frac c{\sin C}=\frac a{\sin A}=2R$$

So, $$(b-c)\sin A=(2R\sin B-2R\sin C)\sin A=2R(\sin A\sin B-\sin C\sin A)$$

or $$(b-c)\sin A=(b-c)\frac a{2R}=\frac{ab-ca}{2R}$$

Utilize the symmetry and use any one of the above derived value to get the desired result.