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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

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Expanding $$A= \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$$ we get $$A=\frac{a^3-a^2 (b+c)-a \left(b^2-3 b c+c^2\right)+(b-c)^2 (b+c)}{a b c}$$ Replace $c$ by $-(a+b)$, expand and simplify.

The result is a whole number.

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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 $\boxed{9}$

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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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Using $c=-a-b$,

$\displaystyle \hspace{.2 in}\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right) \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$

$\displaystyle=\left(\frac{a}{a+2b}-\frac{b}{2a+b}-\frac{a+b}{a-b}\right)\left(\frac{a+2b}{a}-\frac{2a+b}{b}-\frac{a-b}{a+b}\right)$

$\displaystyle=\left(\frac{2(a^2-b^2)}{(a+2b)(2a+b)}-\frac{a+b}{a-b}\right)\left(\frac{2(b^2-a^2)}{ab}-\frac{a-b}{a+b}\right)$

$\displaystyle=(a+b)\left(\frac{2(a-b)}{(a+2b)(2a+b)}-\frac{1}{a-b}\right)(a-b)\left(-\frac{2(a+b)}{ab}-\frac{1}{a+b}\right)$

$\displaystyle=(a+b)\left(\frac{-9ab}{(a+2b)(2a+b)(a-b)}\right)(a-b)\left(-\frac{(a+2b)(2a+b)}{ab(a+b)}\right)=\color{red}{9}$

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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).

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