<s>[INST] Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$ 
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$

$\bf{My\; Try::}$ Given $$\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) = 1+1+1+\frac{b-c}{a}\left(\frac{b}{c-a}+\frac{c}{a-b}\right)+\frac{c-a}{b}\left(\frac{a}{b-c}+\frac{c}{a-b}\right)+\frac{a-b}{c}\left(\frac{a}{b-c}+\frac{b}{c-a}\right)$$
Now How can I Solve after that, Is there is any less complex method  plz explain here , Thanks
 [/INST] Using $c=-a-b$, 
$\displaystyle \hspace{.2 in}\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right) \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\displaystyle=\left(\frac{a}{a+2b}-\frac{b}{2a+b}-\frac{a+b}{a-b}\right)\left(\frac{a+2b}{a}-\frac{2a+b}{b}-\frac{a-b}{a+b}\right)$
$\displaystyle=\left(\frac{2(a^2-b^2)}{(a+2b)(2a+b)}-\frac{a+b}{a-b}\right)\left(\frac{2(b^2-a^2)}{ab}-\frac{a-b}{a+b}\right)$
$\displaystyle=(a+b)\left(\frac{2(a-b)}{(a+2b)(2a+b)}-\frac{1}{a-b}\right)(a-b)\left(-\frac{2(a+b)}{ab}-\frac{1}{a+b}\right)$
$\displaystyle=(a+b)\left(\frac{-9ab}{(a+2b)(2a+b)(a-b)}\right)(a-b)\left(-\frac{(a+2b)(2a+b)}{ab(a+b)}\right)=\color{red}{9}$
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