<s>[INST] 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.
 [/INST] Observe:

This equation is a cubic(degree 3) in "a".So at maximum it can have 3 roots if it is not an identity. But if you show that it has 4 roots , then it becomes an identity.

Roots to try :

b,c(easy ones),0(also easy),-a(think it over).

Now you have shown that it has 4 roots. Thus it becomes an identity.
If b,c are also variables same procedure can be applied for them also.
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