<s>[INST] Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big)+\frac{c-a}{b}\big(\frac{a}{b-c}+\frac{c}{a-b}\big)+\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big)$
Each of these "not yet a number" terms can be expanded to give
$\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big) = \frac{2(c-b)^2}{(c-a)(a-b)}$
$\frac{c-a}{b}\big(\frac{a}{b-c}\big)= \frac{2(a-c)^2}{(c-a)(a-b)}$
$\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big) = \frac{2(b-a)^2}{(b-c)(c-a)}$
Adding each of these terms together and factoring out the two we find that they equal
$2\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=6$
So our total is $9$.
This is problem 160 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom, they give the answer and $9$ is correct. Obviously they don't give the method. 
 [/INST] $$\sum_{cyc}\frac{b-c}{a}=\frac{\sum\limits_{cyc}(a^2b-a^2c)}{abc}=\frac{(a-b)(a-c)(b-c)}{abc}$$ and
$$\sum_{cyc}\frac{a}{b-c}=\frac{\sum\limits_{cyc}a(a-b)(c-a)}{(a-b)(b-c)(c-a)}=$$
$$=\frac{\sum\limits_{cyc}(a^2c-a^3-abc+a^2b)}{-(a-b)(a-c)(b-c)}=\frac{\sum\limits_{cyc}(a^2c+a^2b+abc-a^3+abc-3abc)}{-(a-b)(a-c)(b-c)}=$$
$$=\frac{9abc}{(a-b)(a-c)(b-c)},$$
which gives the answer: $9$.
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