Just a proof of algebra If $a+b+c=0$,
Show that 
$\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this  problem but can't find a solution. Please help me.

Assuming $a,b,c$ are non-zero & distinct
$$F=\dfrac{c}{a-b}\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]$$
$$=1+\frac c{a-b}\cdot\frac{b^2-bc+ca-a^2}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a)-c(b-a)}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a-c)}{ab}$$
$$=1-\frac{c(b+a-c)}{ab}$$
As $b+a=-c,$
$$F=1-\frac{c(-c-c)}{ab}=1+2\cdot\frac{c^2}{ab}$$
Now using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,
 $a+b+c=0\implies a^3+b^3+c^3=3abc$ 
Now, divide both sides by $abc$
