\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}\)
\(\Leftrightarrow\frac{a}{b-c}=\frac{b\left(b-a\right)+c\left(a-c\right)}{\left(a-c\right)\left(b-a\right)}\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ca-c^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)(1)
Tương tự ta cũng có :
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ab-a^2}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}\)(2)
\(\frac{c}{\left(a-b\right)^2}=\frac{a^2-ca+bc-b^2}{\left(c-b\right)\left(a-b\right)\left(a-c\right)}\)(3)
Cộng theo vế (1), (2) và (3) :
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{b^2-ab+ca-c^2+c^2-bc+ab-a^2+a^2-ca+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
\(=\frac{0}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}=0\) ( đpcm )
