prove that:
$\displaystyle a^{2}sin\left(B-C\right) = \left(b^{2} - c^{2}\right)sinA$
where $\displaystyle a,b,c$ are sides of triangle
and $\displaystyle A,B,C$ are the corresponding angles
I'll use the sine and cosine rules.
$\displaystyle a^2\sin(B-C)=a^2(\sin B\cos C-\sin C\cos B)=$
$\displaystyle =a^2\left(\frac{b}{2R}\cdot\frac{a^2+b^2-c^2}{2ab}-\frac{c}{2R}\cdot\frac{a^2+c^2-b^2}{2ac}\right)=$
$\displaystyle =\frac{2a(b^2-c^2)}{4R}=\frac{a}{2R}(b^2-c^2)=(b^2-c^2)\sin A$