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)
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