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I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.

First we just expand and find

$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big)+\frac{c-a}{b}\big(\frac{a}{b-c}+\frac{c}{a-b}\big)+\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big)$

Each of these "not yet a number" terms can be expanded to give

$\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big) = \frac{2(c-b)^2}{(c-a)(a-b)}$

$\frac{c-a}{b}\big(\frac{a}{b-c}\big)= \frac{2(a-c)^2}{(c-a)(a-b)}$

$\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big) = \frac{2(b-a)^2}{(b-c)(c-a)}$

Adding each of these terms together and factoring out the two we find that they equal

$2\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=6$

So our total is $9$.

This is problem 160 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom, they give the answer and $9$ is correct. Obviously they don't give the method.

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4 Answers 4

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plugging $$c=-a-b$$ in the given Expression we get two factors $$\frac{b+a+b}{a}+\frac{-2a-b}{b}+\frac{a-b}{-a-b}=-\frac{(a-b) (2 a+b) (a+2 b)}{a b (a+b)}$$ and the other factor $$\frac{a}{b+a+b}+\frac{b}{-2a-b}+\frac{-a-b}{a-b}=-\frac{9 a b (a+b)}{(a-b) (2 a+b) (a+2 b)}$$ putting Things together we obtain $$9$$ if the denominators are $\ne 0$

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$$\sum_{cyc}\frac{b-c}{a}=\frac{\sum\limits_{cyc}(a^2b-a^2c)}{abc}=\frac{(a-b)(a-c)(b-c)}{abc}$$ and $$\sum_{cyc}\frac{a}{b-c}=\frac{\sum\limits_{cyc}a(a-b)(c-a)}{(a-b)(b-c)(c-a)}=$$ $$=\frac{\sum\limits_{cyc}(a^2c-a^3-abc+a^2b)}{-(a-b)(a-c)(b-c)}=\frac{\sum\limits_{cyc}(a^2c+a^2b+abc-a^3+abc-3abc)}{-(a-b)(a-c)(b-c)}=$$ $$=\frac{9abc}{(a-b)(a-c)(b-c)},$$ which gives the answer: $9$.

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  • $\begingroup$ Dear Michael i think it is $9$ $\endgroup$ Commented Aug 4, 2017 at 18:32
  • $\begingroup$ Yes, I also got this number. $\endgroup$ Commented Aug 4, 2017 at 18:34
  • $\begingroup$ fine then it is all ok $\endgroup$ Commented Aug 4, 2017 at 18:38
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Observe that $(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)=a^3+b^3+c^3-3abc$ so $a^3+b^3+c^3=3abc$. Now \begin{eqnarray*} &&\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)\\ &=&3+\frac{a}{b-c}\left(\frac{c-a}{b}+\frac{a-b}{c}\right)+ \frac{b}{c-a}\left(\frac{a-b}{c}+\frac{b-c}{a}\right)+ \frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\\ &=&3+\frac{a}{b-c}\frac{c^2-ac+bc-b^2}{cb}+.+.\\ &=&3+\frac{a}{b-c}\frac{(a-b-c)(b-c)}{cb}+.+.\\ &=&3+\frac{2(a^3+b^3+c^3)}{abc}=\color{red}{9}. \end{eqnarray*}

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    $\begingroup$ Good point, I didn't want to use the result $a^3+b^3+c^3=3abc$ because that is problem 161, but nice spot. $\endgroup$
    – tr416
    Commented Aug 4, 2017 at 18:50
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It is an olympiad question. Otherwise picking suitable numbers will do: $$a=1,b=2,c=-3$$ $$S=\left(5-2+\frac13 \right)\left(\frac15 -\frac12+3\right)=\frac{10}{3}\cdot \frac{27}{10}=9.$$

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  • $\begingroup$ That's actually a good point, next time I'll plug in some numbers and see what they give, then try to prove that's always the case. Not sure how I didn't see that initially. $\endgroup$
    – tr416
    Commented Aug 5, 2017 at 14:47

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