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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.


The one way I think about this I'll let here :


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.

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I doubt this has a nice proof and even if it has there is probably nothing illuminating about it. – Sebastian Garrido Dec 26 '13 at 23:36
@Anybody Please have a look at my answer also. – user2369284 Dec 27 '13 at 17:15
@Ewin Please read my solution also . – user2369284 Dec 28 '13 at 16:25

Let $x=\frac{a-b}{a+b}$ and so on. We then have: $(1-x)(1-y)(1-z)=(1+x)(1+y)(1+z)$

The terms with an even absolute degree will cancel, therefore: $x+y+z=-xyz$

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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 $f = 0$.

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@Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. – Bill Dubuque Dec 26 '13 at 23:38


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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Could you please elaborate a little more what you have thought ? – Ewin Dec 28 '13 at 17:05
An equation becomes an identity if it is valid for any value of the variable(not just specific values).Consider $x=x$. This is any identity since it is valid for all values of $x(1,3.25,\pi,etc.)$.Had it been an equation(which it is surely not) then at maximum it would have 1 root. But as soon as you show that it has 2 roots, then it becomes an identity.Please reread my answer as I have made some changes to it. – user2369284 Dec 29 '13 at 9:09

I don't know if there is an easier solution but I think that an alternative solution for a,b,c positive real numbers, can be derived using the law of tangents. Of course this will not cover all cases since for a,b,c lengths of sides of a triangle we have $a+b>c$, $b+c>a$, $c+a>b$.

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not all cases, when $a>b+c$, what can you do? – chenbai Dec 27 '13 at 6:23

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