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
Steven Zheng
posted 3 weeks ago
Question
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
\therefore a=c ,or a=b ,or b=c
Steven Zheng
posted 3 weeks ago