Question

Show that

${\cos }^3(\alpha +\theta ){\sin }^3(\beta -\gamma )+{\cos }^3(\beta +\theta ){\sin }^3(\gamma -\alpha )+{\cos }^3(\gamma +\theta ){\sin }^3(\alpha -\beta )$

$=3\cos (\alpha +\theta )\cos (\beta +\theta )\cos (\gamma +\theta )\times \sin (\alpha -\beta )\sin (\beta -\gamma )\sin (\gamma -\alpha )$

Answer 1

Let us put \cos $(\alpha +\theta )\sin (\beta -\gamma )=a$ etc.

$\therefore a+b+c=\sum (\cos \alpha \cos \theta -\sin \alpha \sin \theta )\sin (\beta -\gamma )$

$=\cos \theta \sum \cos \alpha \sin (\beta -\gamma )-\sin \theta \sum \sin \alpha \sin (\beta -\gamma )=0$

It is easy to observe on expansion that both the above sigmas are zero.Hence$a+b+c=0$ that from Algebra we know that

$a^3+b^3+c^3=3abc.$

On putting the values of a, b, c, we prove the given result.