$\sin^3 a\sin(b-c)+\sin^3b\sin(c-a)+\sin^3c\sin(a-b)+\sin(a+b+c)\sin(b-c)\sin(c-a)\sin(a-b)=0$ 
Verify that$$\sin^3 a\sin(b-c)+\sin^3b\sin(c-a)+\sin^3c\sin(a-b)$$
  $$+\sin(a+b+c)\sin(b-c)\sin(c-a)\sin(a-b)=0.$$

I tried to break $\sin^3a$ into $\sin^2\cdot\sin a$ and use it to make sum with $\sin(b-c)$, but eventually messed up everything. Please help.

$$\sum_{cyc}\sin^3a\sin(b-c)=\frac{1}{4}\sum_{cyc}(3\sin{a}-\sin3a)\sin(b-c)=$$
$$=\frac{1}{8}\sum_{cyc}\left(3\cos(a-b+c)-3\cos(a+b-c)-\cos(3a-b+c)+\cos(3a+b-c)\right).$$
$$\sin(a+b+c)\prod_{cyc}\sin(a-b)=\frac{1}{2}\sin(a+b+c)\sin(b-c)(\cos(2a-b-c)-\cos(b-c))=$$
$$=\frac{1}{4}\sin(a+b+c)(-\sin(2c-2a)-\sin(2a-2b)-\sin(2b-2c))=$$
$$=-\frac{1}{4}\sin(a+b+c)\sum_{cyc}\sin(2b-2c)=-\frac{1}{8}\sum_{cyc}(\cos(3c+a-b)-\cos(3b+a-c))=$$
$$=\frac{1}{8}\sum_{cyc}(\cos(3a+c-b)-\cos(3a+b-c))$$
and since 
$$\sum\limits_{cyc}(\cos(a-b+c)-\cos(a+b-c))=\sum\limits_{cyc}(\cos(b-c+a)-\cos(a+b-c))=0,$$ 
we are done!
