<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] 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 $(4)$:
$$\underbrace{x^2+y^2+z^2}_{3}+2(\underbrace{xy+yz+zx}_{0})=9x^2y^2z^2 \Rightarrow 3=9x^2y^2z^2 \Rightarrow xyz=\pm \frac1{\sqrt{3}}.$$
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