For $a$, $b$, $c$ the angles of a right triangle, show that $\left(\sum_{cyc}\sin a\sin b\sin(a-b)\right)+\sin(a-b)\sin(b-c)\sin(c-a)=0$ 
Given that $a$, $b$, $c$ are the angles of a right-angled triangle, prove that:
  $$\begin{align}
\sin a\sin b\sin(a-b) &+\sin b\sin c\sin(b-c)+\sin c\sin a\sin(c-a) \\ &+\sin(a-b)\sin(b-c)\sin(c-a)=0
\end{align}$$

I know I'm supposed to use the properties of polynomials for this, as this question was found on the chapter on polynomials. I've tried having these values be roots of some function but don't know how to carry it out.
I also know that I can consider one of the angles to be $90^\circ$, so the sine of that would be $1$, but that dosen't really simplify it too much.
 A: $$4\sin A\sin B\sin(A-B)$$
$$=2\sin(A-B)[\cos(A-B)-\cos(A+B)]$$
$$=\sin(2A-2B)-(\sin2A-\sin2B)$$
Now let $2A-2B=2x$ etc. $\implies x+y+z=0$
$$\sin2x+\sin2y+\sin2z$$
$$=2\sin x\cos x+2\sin(y+z)\cos(y-z)$$
$$=2\sin x\cos(-y-z)+2\sin(-x)\cos(y-z)$$
$$=2\sin x\cos(y+z)-2\sin x\cos(y-z)$$
$$=-2\sin x[\cos(y-z)-\cos(y+z)]=?$$
A: If $c$ is your right angle then $a+b=90^{\circ}$.
Then $\sin (c-a) = \sin b$ and so forth.
This alone (along with $\sin c = 1$) will get you to the answer.
Care to try?
Spoiler:

For the last term, $\sin (a-b) \sin (b-c) \sin (c-a) = [\sin (a-b)](-\sin a) [\sin b ]$, which cancels the first term. The middle two terms become $[\sin b] (1) (-\sin a) + (1)\sin a \sin b$, which add to zero.

A: Let $a$ be a right angle without loss of generality. Then $c=90^{°}-b$.
$\sin a \sin b \sin (a - b) + \sin b \sin c \sin(b - c) + \sin c \sin a \sin (c -a ) + \sin(a - b)\sin(b - c)\sin(c - a)$
$=\sin 90^{°} \sin b \sin (90^{°} - b) + \sin b \sin (90^{°}-b) \sin(b - (90^{°}-b)) + \sin (90^{°}-b) \sin 90^{°} \sin ((90^{°}-b) - 90^{°}) + \sin(90^{°} - b)\sin(b - (90^{°}-b))\sin((90^{°}-b) - 90^{°})$
Using the results $\sin(90^{°})=1$ and $\sin(90^{°}-b)=\cos b$
$=\sin b \cos b+ \sin b \cos b \sin(2b - 90^{°}) - \cos b \sin b - \cos(b)\sin(2b - 90^{°})\sin b$
$=0$
A: The identity $\,\text{id}_{3,4,1,3a}\,$ in my list of
Special Algebraic Identities is given by
$$ 0=a\,b\,(a-b)+b\,c\,(b-c)+c\,a\,(c-a)+(a-b)(b-c)(c-a). \tag{1}$$
This algebraic identity is verified by expanding out the products.
The related functional equation
$$ 0 \!=\! f(a)f(b)f(a\!-\!b)\!+\!f(b)f(c)f(b\!-\!c)\!+\!
 f(c)f(a)f(c\!-\!a)\!+\!f(a\!-\!b)f(b\!-\!c)f(c\!-\!a) \tag{2} $$
is satisfied by Jacobi elliptic functions of which limiting cases are the 
trigonometric $\sin$ and $\sinh$ functions. Using the equality $\,\sin(x)=(\exp(ix)-\exp(-ix))/2\,$ and expanding and simplifying, the identity $(2)$
for $\sin$ can be verified. No restriction on $\,a,b,c\,$ is needed.
Note that we can write $\,\sin(a) = (X-1/X)/2\,$ where $\,X:=\exp(ia)\,$
and also $\,\sin(a-b) = (X/Y-Y/X)/2\,$ where $\,Y:=\exp(ib).$ Of course,
in your case, if $\,c=90^\circ\,$ then certain simplifications hold. For
example, $\,\sin(c)=1,\,$ and $\,\sin(c-a)=\cos(a)\,$. You will
also need to use the subtraction equation
$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b).$
