Question

Prove that

${\sin }^3(\beta -\gamma ){\sin }^3(\alpha -\delta )+{\sin }^3(\gamma -\alpha ){\sin }^3(\beta -\delta )+{\sin }^3(\alpha -\beta ){\sin }^3(\gamma -\delta )$
$=3\sin (\alpha -\beta )\sin (\beta -\gamma )\sin (\gamma -\delta )\sin (\alpha -\delta )\sin (\beta -\delta )\sin (\gamma -\delta )$

Answer 1

Let $a=\sin (\beta -\gamma )\sin (\alpha -\delta ),b=......,c=....$
$a=\frac{1}{2}\left[\cos \right(\beta -\gamma -\alpha +\delta )-\cos (\beta -\gamma +\alpha -\delta \left)\right]$
It is easy to observe that $a+b+c=0$
$\therefore a^3+b^3+c^3=3abc,$ etc