<s>[INST] 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.
 [/INST] 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
$$xyz(x+y+z)=0$$
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