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.
 A: Given: $x-\frac1x=y,y-\frac1y=z,z-\frac1z=x$, you found correctly:
$$x-\frac1x+y-\frac1y+z-\frac1z=y+z+x \Rightarrow \frac1x+\frac1y+\frac1z=0 \Rightarrow xy+yz+zx=0 \ \ (1)\\
x^2-1+y^2-1+z^2-1=xy+yz+zx \Rightarrow x^2+y^2+z^2=3 \ \ (2)$$
Square each and add them all:
$$x^2+\frac1{x^2}-2+y^2+\frac1{y^2}-2+z^2+\frac1{z^2}-2=y^2+z^2+x^2 \Rightarrow \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=6 \Rightarrow \\
x^2y^2+y^2z^2+z^2x^2=6x^2y^2z^2 \ \ (3)$$
Square $(1)$:
$$\underbrace{x^2y^2+y^2z^2+z^2x^2}_{6x^2y^2z^2}+2xyz(x+y+z)=0 \Rightarrow x+y+z=-3xyz \ \ (4)$$
Square $(4)$:
$$\underbrace{x^2+y^2+z^2}_{3}+2(\underbrace{xy+yz+zx}_{0})=9x^2y^2z^2 \Rightarrow 3=9x^2y^2z^2 \Rightarrow xyz=\pm \frac1{\sqrt{3}}.$$
A: Hint: By the equation #(3) we get in #(1)
$$z-1/z-\frac{1}{z-\frac{1}{z}}=y$$
and with $z=y-\frac{1}{y}$ we get
$$3y^6-9y^4+6y^2-1=0$$
Multiplying all terms together gives
$$x^2(1-z^2)+y^2(1-z^2)+z^2(1-y^2)=0$$
substituting
$$xy=x^2-1$$
$$yz=y^2-1$$
$$xz=z^2-1$$
so we get
$$xyz(x+y+z)=0$$
A: 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
A: Eliminating $x$ and $z$ between the equations we obtain:
$$3y^6-9y^4+6y^2-1=0.$$
By symmetry we know that $x$ and $z$ must also satisfy this equation. Hence $x^2$, $y^2$ and $z^2$ are roots of $$3y^3-9y^2+6y-1=0.$$
Therefore $(xyz)^2=\frac{1}{3}$ and hence $xyz=\pm\frac{1}{\sqrt 3}$.
In addition I think this may be one of the simplest direct solutions:
Adding the equations: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>xy+yz+zx=0.$$ Clearing the denominators of the original equations ($x^2-xy-1=0, \dots$) and adding we obtain $$x^2+y^2+z^2=xy+yz+zx+3=3$$ Hence $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=3=>(x+y+z)=\pm\sqrt{3}$$Rearranging the original equations ($y+\frac{1}{x}=x,\dots$), multiplying them and simplifying: $$(y+\frac{1}{x})(z+\frac{1}{y})(x+\frac{1}{z})=xyz =>(x+y+z)=-\frac{1}{xyz}.$$ Combining we obtain the result $$xyz=\pm\frac{1}{\sqrt{3}}.$$
