If $$a^2=b+c, b^2=c+a, c^2=a+b$$, then the value of $$3(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$$

Given : $$a^2=b+c$$

Adding $$'a'$$ on both sides

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

=> $$a(a+1)=a+b+c$$

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

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

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

To find : $$3(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$$

Substituting values from equations (i),(ii) and (iii)

= $$3(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c})$$

= $$3(\frac{a+b+c}{a+b+c})=3$$

=> Ans - (C)