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

I think your method is efficient and pretty neat.
Continuing from where you left.
Observe that 
$$\dfrac{b-c}{a}\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)=\dfrac{b-c}{a}\left(\dfrac{b(a-b)+c(c-a)}{(a-b)(c-a)}\right)=\dfrac{(b-c)^2 \cdot (a-b-c)}{a(a-b)(c-a)}$$
Now, using $a+b+c=0$ we get,
$$\dfrac{b-c}{a}\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)=\dfrac{2(b-c)^3}{(a-b)(b-c)(c-a)}$$
Similarly, other terms are dealt with.
Now, substitute $x=b-c,y=c-a$ and $z=a-b$
And then use $x^3+y^3+z^3=3xyz$ as $x+y+z=0$ 

Finally, we get the answer as $\boxed{9}$

