How to prove $\sin{b}\sin{c}\sin{(b-c)}(\sin^2{b}+\sin^2{c}+\sin^2{(b-c)})+\dots=0$ 
If $a,b,c\in (0,\pi)$ and $a+b+c=\pi$, show that:
  $$\sin b \sin c \sin(b-c) \left(\sin^2 b + \sin^2 c + \sin^2(b-c)\right) \\ 
+ \sin c \sin a \sin(c-a) \left(\sin^2 c + \sin^2 a + \sin^2(c-a)\right) \\ 
+ \sin a \sin b \sin(a-b) \left(\sin^2 a + \sin^2 b + \sin^2(a-b)\right) \\
+ \sin(b-c)\sin(c-a)\sin(a-b) \left(\sin^2(b-c) + \sin^2(c-a) + \sin^2(a-b)\right) = 0$$

This problem is from my friend. He asked me and I can't solve it. Thank you for your help.
My idea: let $\sin{a}=x,\sin{b}=y,\sin{c}=z$. Then $LHS=\cdots$.
I fell very hard, so I can't do any work. Thank you.
 A: Consider the following configuration:

where $ABC$ is an acute triangle and $X'$ is the symmetric of $X$ with respect to the perpendicular bisector of $YZ$. The six depicted points are clearly concyclic, and by assuming that the diameter of the circumcircle is one we have:
$$ z = XY = \sin\widehat{Z},\quad XX'=\left|\sin(\widehat{Y}-\widehat{Z})\right|.$$
Without loss of generality, we can further assume that $\widehat{A}\geq\widehat{B}\geq\widehat{C}$, as depicted above. By applying the Ptolemy's theorem to the isosceles trapezoid $AA'CB$ we get:
$$ AA'\cdot a = AA'\cdot BC = AC^2 - AB^2 = b^2-c^2,$$
in the same way:
$$ BB'\cdot b = BB'\cdot AC = BC^2 - AB^2 = a^2-c^2,$$
$$ CC'\cdot c = CC'\cdot AB = BC^2 - AC^2 = a^2-b^2.$$
Now we have to prove:
$$ bc AA'(b^2+c^2+AA'^2) - ac BB'(a^2+c^2+BB'^2) + ab CC' (a^2+b^2+CC'^2) - AA'BB'CC'(AA'^2+BB'^2+CC'^2)=0.$$
We multiply everything by $a^3 b^3 c^3$ in order to remove $AA',BB'$ and $CC'$ through the previous Ptolemy's identities. We end with:
$$\begin{eqnarray*} 
  &b^4 c^4 a AA'(a^2b^2+a^2c^2+a^2 AA'^2)\\
  -&a^4 c^4 b BB'(a^2b^2+b^2c^2+b^2 BB'^2)\\
  +&a^4 b^4 c CC'(a^2c^2+b^2c^2+c^2 CC'^2)\\
  -&a AA' b BB' c CC'(b^2c^2 a^2 AA'^2+a^2 c^2 b^2 BB'^2+a^2 b^2 + c^2 CC'^2)=0,
  \end{eqnarray*}$$
or:
$$\sum_{cyc}b^4 c^4(b^2-c^2)(a^2b^2+a^2c^2+(b^2-c^2)^2)+\left(\prod_{cyc}(b^2-c^2)\right)\cdot\sum_{cyc}b^2c^2(b^2-c^2)^2 = 0.$$
$$\sum_{cyc}a^2 b^4 c^4(b^4-c^4)+ \sum_{cyc}b^4 c^4(b^2-c^2)^3+\left(\prod_{cyc}(b^2-c^2)\right)\cdot\sum_{cyc}b^2c^2(b^2-c^2)^2 = 0.$$
Since $\prod_{cyc}(x-y)=-\sum_{cyc}xy(x-y)$, the last identity is tedious to check but straightforward.
