In triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $\frac{a^{2} \sin (B-C)}{\sin A}+\frac{b^{2} \sin (C-A)}{\sin B}+\frac{c^{2} \sin (A-B)}{\sin C}=0$.

$\mathrm{LHS}=\frac{a^{2} \sin (B-C)}{\sin A}+\frac{b^{2} \sin (C-A)}{\sin B}+\frac{c^{2} \sin (A-B)}{\sin C}$

$=\frac{k^{2} \sin ^{2} A \sin (B-C)}{\sin A}+\frac{k^{2} \sin ^{2} B \sin (C-A)}{\sin B}+\frac{k^{2} \sin ^{2} C \sin (A-B)}{\sin C}$

$=k^{2} \sin A \sin (B-C)+k^{2} \sin B \sin (C-A)+k^{2} \sin C \sin (A-B)$

$=k^{2}[\sin A(\sin B \cos C-\sin C \cos B)+\sin B(\sin C \cos A-\sin A \cos C)+\sin C(\sin A \cos B-\sin B \cos A)]$

$=k^{2}(\sin A \sin B \cos C-\sin A \sin C \cos B+\sin B \sin C \cos A-\sin A \sin B \cos C+\sin A \sin C \cos B-\sin C \sin B \cos A)$

$=0=\mathrm{RHS}$

Hence proved.