Question

In any triangle $ABC,a^3\cdot \sin (B-C)+b^3\cdot \sin (C-A)+c^3\cdot \sin (A-B)=?$

Answer 1

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

Now, $a^3\sin (B-C)+b^3\sin (C-A)+c^3\sin (A-B)$

$=a^3\left(\frac{b^2-c^2}{a^2}\right)+b^3\left(\frac{c^2-a^2}{b^2}\right)+c^3\left(\frac{a^2-b^2}{c^2}\right)$

$={ab}^2-{ac}^2+{bc}^2-{ba}^2+{ca}^2-{cb}^2$

$=ab(b-a)+bc(c-b)+ac(a-c)$