<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] We have $$A = \left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\frac{bc(b-c) + ca(c-a) + ab(a-b)}{abc} = \frac{-3abc - a^3 -b^3 -c^3+ a^2(c-b) + b^2(a-c) +c^2(b-a)}{abc}.$$
Note that $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$ or $a^3+b^3+c^3 = 3abc$.
Then $$A = \frac{(a-b)(c-b)(a-c)}{abc}$$
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