Calculate value of expression.

[jagdish]

If $a+b+c=0$.then find value of expression $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right).\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)=$

[Zzzz]

\[\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right).\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) \]
\[= \frac {a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)}{(a-b)(b-c)(c-a)}\cdot \frac {bc(b-c)+ca(c-a)+ab(a-b)}{abc}\]
\[= \frac {a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)+ 9abc - 9abc}{(a-b)(b-c)(c-a) } \cdot \frac {-(a-b)(b-c)(c-a)}{abc}\]
\[ = - \frac {a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)+ 9abc - 9abc}{abc}\]

Let, $f(a)= a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)+ 9abc$
$\therefore f(-b-c)=0$. That means $(a+b+c)$ divides $f(a)$.
Let, $f(a)=k(a+b+c) \Rightarrow f(a)=k\cdot 0=0$

Now,
\[ = - \frac {a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)+ 9abc - 9abc}{abc}\]
\[ = - \frac {f(a) -9abc}{abc}\]
\[ = - \frac {0-9abc}{abc} \]
\[ = 9 \]