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

We have from GAVD's computation that
$$
\begin{align}
A &:= \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)
\\
&= \left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\left(\frac{bc(b-c) + ca(c-a) + ab(a-b)}{abc}\right)
\\&=\left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\left(-\frac{(a-b)(b-c)(c-a)}{abc}\right)
\\
&= \frac{a(a-b)(a-c)+b(b-c)(c-a)+c(c-a)(c-b)}{abc}
\\
&=\frac{a^3+b^3+c^3-a^2(b+c)-b^2(c+a)-c^2(a+b)+3abc}{abc}\,.
\end{align}$$
Since $a+b+c=0$, we have
$$A=\frac{2\left(a^3+b^3+c^3\right)+3abc}{abc}=\frac{2\left(a^3+b^3+c^3-3abc\right)}{abc}+9\,.$$
As $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-bc-ca-ab\right)=0$, we obtain $A=9$ (given that $a,b,c\neq 0$ are pairwise distinct).
