<s>[INST] Just a proof of algebra If $a+b+c=0$,
Show that 
$\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this  problem but can't find a solution. Please help me.
 [/INST] If you expand the expression and factor  you get $$\frac{(a^3+b^3+c^3-ab^2-a^2b-ac^2-a^2c-bc^2-b^2c+3abc)(b-c)(c-a)(a-b)}{abc (b-c)(c-a)(a-b)}$$ and if you then use $c=-a-b$ you get $$\frac{-9a^2b-9ab^2}{-a^2b-ab^2}=9$$ providing you have not divided by zero (so the $a,b,c$ must be distinct and non-zero as  Suzu Hirose said in the comments). 
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