Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If 
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$ 
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$  .
I am stuck here please help.
This image gives some hint, but I am unable to understand it.

Note that adding up the three equations gives you:
$$x+y+z =x+y+z - \frac{1}{x} - \frac{1}{y} - \frac{1}{z}$$
, so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$
Try it from there
