If $$\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$$ then find the value of $$\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$$

Given : $$\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$$

=> $$\frac{a^2}{b+c}=1$$

=> $$a^2=b+c$$

Adding $$'a'$$ on both sides, => $$a(a+1)=a+b+c$$

=> $$\frac{1}{a+1}=\frac{a}{a+b+c}$$ -----------(i)

Similarly, $$\frac{1}{b+1}=\frac{b}{a+b+c}$$ --------(ii)

and $$\frac{1}{c+a}=\frac{c}{a+b+c}$$ -------------(iii)

To find : $$\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$$

= $$2(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$$

Substituting values from equations (i),(ii) and (iii), we get :

= $$2(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c})$$

= $$2 \times (\frac{a+b+c}{a+b+c})=2$$

=> Ans - (C)