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Question 128
If $$x^2=y + z, y^2=z + x, z^2=x+y$$, then the value $$\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$$
Solution
$$x^2 = y+z$$
$$y^2 = z+x$$
on substracting above two eq. we will get
$$x^2 - y^2 = y - x$$
So either $$x=y$$Â
or $$x+y= -1$$ (it is not possible as $$z^2$$ can not be negative)
So $$x=y=z=2$$
So given eq. will reduce to a value 1
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