问题标题：
一条分式数学题,希望能有高手为我详细的解答.已知:a+b+c=0且(b-c/a)+(c-a/b)+(a-b/c)=0,求证:(bc+b-c/b^2c^2)+(ca+c-a/c^2a^2)+(ab+a-b/a^2b^2)=0.
问题描述：
一条分式数学题,希望能有高手为我详细的解答.
已知:a+b+c=0且(b-c/a)+(c-a/b)+(a-b/c)=0,求证:(bc+b-c/b^2c^2)+(ca+c-a/c^2a^2)+(ab+a-b/a^2b^2)=0.
满春涛回答：
(bc+b-c)/(b^2c^2)+(ca+c-a)/(c^2a^2)+(ab+a-b)/(a^2b^2)
=a^2(bc+b-c)/(b^2c^2a^2)+b^2(ca+c-a)/(c^2a^2b^2)+c^2(ab+a-b-1)/(a^2b^2c^2)
=(a^2bc+a^2b-a^2c+b^2ac+b^2c-b^2a+c^2ab+c^aa-c^2b)/(a^2b^2c^2)
=[(a+b+c)abc+a^2b-a^2c+b^2c-b^2a+c^2a-c^2b)]/(a^2b^2c^2)
=[(a+b+c)abc+a^2b-b^2a+a^2c-c^2a+b^2c-c^2b)]/(a^2b^2c^2)
=[(a+b+c)abc+(a-b)ab+(b-c)bc+(c-a)ac]/(a^2b^2c^2)
={(a+b+c)abc+abc[(b-c)/a+(c-a)/b+(a-b)/c]}/(a^2b^2c^2)
=0
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