Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I wondering how to show easily this identity ? As a matter of the fact, it's a beautiful identity.
$$-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}$$
The one way I think about this I'll let here :
$$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b)(b+c)(c+a)+(b-c)(c+a)(a+b)+(c-a)(c+b)(b+a)}{(a+b)(b+c)(c+a)}$$
Now, I'll take one of the term of RHS
$(a −b)(b +c)(c + a)\\ =(a −b)(b−c + 2c)(a −c + 2c)\\ =(a −b)(b −c)(a −c)+ 2c(a −b)(b−c + a −c + 2c) \\=(a −b)(b −c)(a −c)+ 2c(a −b)(a +b)$
Similarly, We'll do that with the remainder
$(b−c)(a+b)(c + a)+(c − a)(a +b)(b+ c)\\=(a +b)[(b −c)(c + a)+(c −a)(b +c)]\\=(a +b)(bc +ba −c^{2} −ca +cb+c^{2} −ab −ac)\\= (a +b)(2bc − 2ac)\\= −2c(a −b)(a +b)$
and we get :
$$\boxed{\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}}$$
So it's too many work to show this identity. I just want to know if there's a simple way to show that or I don't know. I'm questing that because I didn't find anything on internet.

Bringing all to one side and clearing denominators, we wish to prove zero
$$f(c) = (a\!-\!b)(b\!-\!c)(c\!-\!a)+(a\!-\!b)(b\!+\!c)(c\!+\!a)+(b\!-\!c)(a\!+\!b)(c\!+\!a) + (c\!-\!a)(a\!+\!b)(b\!+\!c)$$
Note $\ f(a) = (a\!-\!b)(b\!+\!a)2a\!+\!(b\!-\!a)(a\!+\!b)2a = 0.\ $  By symmetry $f(b) = 0.$  
The coefficent of $c^2$ is $\ {-}(a\!-\!b) \!+\! (a\!-\!b) -(a\!+\!b)\!+\!(a\!+\!b) = 0.$ 
Thus $f(c)$ is a polynomial in $\,c\,$ of degree $\le 1$ with $\,2\,$ roots, so $f = 0$.
