Question on solving one algebra relation to obtain another If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:
 If $(1-2x)(1-2y)(1-2z)=-27$, then evaluate $x+y+z$ where $x=\frac{b}{a+b}, \ y=\frac{c}{b+c}, \ z=\frac{a}{c+a}$
 I'm not sure how to solve this. I think it may use inequalities or pure algebra. 
Please help me out here. Thanks a lot :)

Let's call $\begin{cases} U=\frac a{a+b}+\frac b{b+c}+\frac c{c+a}\\V=\frac b{a+b}+\frac c{b+c}+\frac a{c+a}\end{cases}$
We have trivially $U+V=3$
Now $U-V=\frac {a-b}{a+b}+\frac {b-c}{b+c}+\frac {c-a}{c+a}=\cdots=-\frac {a-b}{a+b} \times\frac {b-c}{b+c}\times\frac {c-a}{c+a}=27$
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$
Therefore $U=15$ and $V=-12$.
