If $$x^2 = y + z , y^2 = z + x$$ and $$z^2 = x + y$$ then the value of $$\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}$$ is

Given, $$x^2 = y + z$$

So, $$x^2+x=x+y+z$$

or, $$x\left(x+1\right)=x+y+z$$

or, $$\dfrac{1}{x+1}=\dfrac{x}{x+y+z}$$ --->(1)

Similarly, $$y^2 = x + z$$

Adding, $$y$$ to both sides,

$$y^2+y=y+x+z$$

or, $$y\left(y+1\right)=x+y+z$$

or, $$\dfrac{1}{y+1}=\dfrac{y}{x+y+z}$$ --->(2)

Similarly, $$z^2 = y + z$$

So, $$\dfrac{1}{z+1}=\dfrac{z}{x+y+z}$$ --->(3)

Now, adding (1), (2) and (3),

$$\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=\dfrac{x+y+z}{x+y+z}=1$$

So, correct answer is option (B).