If $$ a+b+c=0 $$ , $$ \dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}=0 $$ , prove $$ \dfrac{bc+b-c}{b^2c^2}+\dfrac{ca+c-a}{c^2a^2}+\dfrac{ab+a-b}{a^2b^2}=0 $$
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If $$ a+b+c=0 $$ , $$ \dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}=0 $$ , prove $$ \dfrac{bc+b-c}{b^2c^2}+\dfrac{ca+c-a}{c^2a^2}+\dfrac{ab+a-b}{a^2b^2}=0 $$  
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Uniteasy Admin · Accepted Answer

$$ \dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c} $$
$$ =\dfrac{bc(b-c)+ac(c-a)+ab(a-b)}{abc} $$
$$ =\dfrac{b^2c-bc^2+ac^2-a^2c+a^2b-ab^2}{abc} $$
$$ =\dfrac{b^2(c-a)-b(c^2-a^2)+ac(c-a)}{abc} $$
$$ =\dfrac{(c-a)(b^2-bc-ab+ac)}{abc} $$
$$ =\dfrac{(c-a)[b(b-a)-c(b-a)]}{abc} $$
$$ =\dfrac{(c-a)(b-a)(b-c)}{abc} $$
$$ =0 $$
$$ 	herefore a=c $$ ,or $$ a=b $$ ,or $$ b=c $$

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