Solution to Problem #433: Proving a+b+c=0 yields 9

[anemone]

Here is this week's POTW:

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Show that if $a+b+c=0$, then $\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right)=9$.

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[anemone]

Congratulations to kaliprasad for his correct solution!(Cool)

You can find the suggested solution as below:

Spoiler: Solution of other

Let $x=b-c,\,y=c-a,\,z=a-b$. Since $a+b+c=0$, we have

$y-z=b+c-2a=-3a$, by the same token we get $z-x=-3b$ and $x-y=-3c$.

By using the identity $\dfrac{y-z}{x}+\dfrac{z-x}{y}+\dfrac{x-y}{z}=-\dfrac{(y-z)(z-x)(x-y)}{xyz}$ for $x,\,y,\,z \ne 0$, and replacing the variables $x,\,y$ and $z$ with $a,\,b$ and $c$, we get

$\dfrac{-3a}{b-c}+\dfrac{-3b}{c-a}+\dfrac{-3c}{a-b}=-\dfrac{(-3a)(-3b)(-3c)}{(b-c)(c-a)(a-b)}$, which simplifies to

$\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=-\dfrac{9abc}{(b-c)(c-a)(a-b)}$--(1)

Now, if we set $x=a,\,y=b$ and $z=c$ in the above identity, we get

$\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}=-\dfrac{(b-c)(c-a)(a-b)}{abc}$--(2)

Multiplying the last two equalities (1) and (2), we get the desired result.