Question

$x^2=y+z;y^2=x+z;z^2=x+y$
$\frac{1}{(x+1)}+\frac{1}{(y+1)}+\frac{1}{(z+1)}$

Answer 1

$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$
Multiply both the numerator and denominator of the first term by x, second term by y, third term by z. Doing that we get:
$\frac{x}{x+x^2}+\frac{y}{y+y^2}+\frac{z}{z+z^2}$
Now consider $x^2=y+z$
Adding x on both sides of the equation, we get:
$x+x^2=x+y+z$
Taking the reciprocal and multyplying by x on both sides:
$\frac{x}{x+x^2}=\frac{x}{(x+y+z)}$
Doing the same for the other terms we get:
$\frac{y}{x+y+z}and\frac{z}{(x+y+z)}$
Adding all three terms we get: $\frac{(x+y+z)}{(x+y+z)}=1$
So in the case when x,y,z not equal to 0 we get 1
If we consider the case when one of the terms can be 0, then the answer would be e
so the answer to this question will be either 1 or 3