\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