$\dfrac{4}{a-b}.(\dfrac{1}{b-c}+\dfrac{1}{c-a})=\dfrac{4}{(b-c).(a-c)}$
$\dfrac{4}{a-b}.(\dfrac{1}{b-c} + \dfrac{1}{c-a})=\dfrac{4}{(b-c).(a-c)}$