An algebra problem by Sourasish Karmakar

$\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} = 1\\ \color{#3D99F6}{\text{Multiply both sides by : }} \color{#D61F06}{(a+b+c)} \\ \color{#D61F06}{(a+b+c)}[\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b}] = 1\color{#D61F06}{(a+b+c)} \\ \dfrac{a\color{#D61F06}{(a+b+c)}}{b+c} + \dfrac{b\color{#D61F06}{(a+b+c)}}{c+a} + \dfrac{c\color{#D61F06}{(a+b+c)}}{a+b} = (a+b+c) \\ a(\dfrac{a+(\color{#3D99F6}{b+c})}{b+c}) + b(\dfrac{(b+(\color{#3D99F6}{c+a})}{c+a})) + c(\dfrac{(c+(\color{#3D99F6}{a+b})}{a+b})) = (a+b+c) \\ a(\dfrac{a}{b+c} + (\color{#D61F06}{\dfrac{b+c}{b+c}})) + b(\dfrac{b}{c+a} + (\color{#D61F06}{\dfrac{c+a}{c+a}})) + c(\dfrac{c}{a+b} + (\color{#D61F06}{\dfrac{a+b}{a+b}})) = (a+b+c) \\ a(\dfrac{a}{b+c} + 1) + b(\dfrac{b}{c+a} + 1) + c(\dfrac{c}{a+b} + 1) = (a+b+c) \\ \dfrac{a^2}{b+c} + \color{#3D99F6}{a} + \dfrac{b^2}{c+a} + \color{#3D99F6}{b} + \dfrac{c^2}{a+b} + \color{#3D99F6}{c} = \color{#D61F06}{(a+b+c)} \\ \dfrac{a^2}{b+c} + \dfrac{b^2}{c+a} + \dfrac{c^2}{a+b} + \color{#D61F06}{(a+b+c)} = \color{#D61F06}{(a+b+c)} \\ \dfrac{a^2}{b+c} + \dfrac{b^2}{c+a} + \dfrac{c^2}{a+b} = 0$

Note : You can also do this process in reverse, from last step to first step :)