Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$. This is the final step of one old competition problem, and in textbook it simply says "reader can prove this", but I found this a bit nontrivial. 

Show that  $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$.

I tried a lot of things and formula on this, but never succeeded.
Can someone help? 

$LHS=\sin^2 20^\circ \sin 40^\circ =\sin 20^\circ\cdot\frac{1}{2}(\cos20^\circ -\cos60^\circ)$
$=\frac{1}{2}(\sin 20^\circ \cos20^ \circ-\frac{1}{2}\sin 20^\circ)=\frac{1}{4}(\sin 40^\circ-\sin 20^\circ)$
$ =\frac{1}{4}\cdot2\sin 10^\circ\cos 30^\circ=\frac14\sqrt3\sin 10^\circ. $
$RHS=\sin10^\circ \sin30^\circ \sin60^\circ=\frac14\sqrt3\sin 10^\circ.$
$LHS=RHS.$
