<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] Observe that $(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)=a^3+b^3+c^3-3abc$ so $a^3+b^3+c^3=3abc$. Now
\begin{eqnarray*}
&&\left( \frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}  \right) \left( \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}  \right)\\
&=&3+\frac{a}{b-c}\left(\frac{c-a}{b}+\frac{a-b}{c}\right)+
\frac{b}{c-a}\left(\frac{a-b}{c}+\frac{b-c}{a}\right)+
\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\\
&=&3+\frac{a}{b-c}\frac{c^2-ac+bc-b^2}{cb}+.+.\\
&=&3+\frac{a}{b-c}\frac{(a-b-c)(b-c)}{cb}+.+.\\
&=&3+\frac{2(a^3+b^3+c^3)}{abc}=\color{red}{9}.
\end{eqnarray*}
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