Finding $x+y+z$ If $x+1/x = y$, $y+1/y=z$, $z+1/z=x$, then find $x+y+z$. Is there any way to do so without taking out the values of $x$, $y$, $z$?
Please help,
 A: Adding the three equations, you get $\frac1x + \frac1y + \frac1z=0$ or $xy + yz + zx = 0$.
The equations can be written also as $x^2+1 = xy, y^2+1 = yz, z^2+1 = zx$, so $x^2+y^2+z^2 = -3$ which implies there are no solutions possible among real numbers.
Not sure if that qualifies as your answer as you didn't want to solve for $x, y, z$ first...
A: Let us first write down the equations:
$$x+\frac{1}{x}=y, \; y + \frac{1}{y}=z,\; z+\frac{1}{z}=x$$  
We now add these equations together: 
$$x+\frac{1}{x}+ y + \frac{1}{y}+z+\frac{1}{z}=x +y+z$$
So we can see cancelling out $x+y+z$ gives us something, which was our motivation in adding them up:
$$\frac{1}{x} + \frac{1}{y}+\frac{1}{z}=0$$
Expanding, we get $xy+yz+xz=0$. 
Here, we can note the identity $(x+y+z)^2 = x^2 +y^2+z^2 + 2(xy+yz+zx)$. Hence, if we can find $x^2+y^2+z^2$, we are done.
This can be done by realizing that by multiplying $x$ on both sides of $x+\frac{1}{x} = y$, gives us $x^2+1=xy$. Doing this with other equations and adding them all, we have:
$$x^2+1+y^2+1+z^2+1=xy+yz+xz$$
Substituing $xy+yz+xz=0$, we have $x^2 +y^2+z^2 = -3$. So,
$$x+y+z=\pm\sqrt{x^2 +y^2+z^2 + 2(xy+yz+zx)} = \pm\sqrt{-3+2(0)} = \pm\sqrt{-3} = \pm\sqrt{3}i$$
