<s>[INST] What is the value of $\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}$? Let $x,y,z$ are real numbers such that $x,y,z \geq -1$, $xyz\not= 0$ and
\begin{matrix} \sqrt{x+1} + \sqrt{y+2} +\sqrt{z+3}= \sqrt{y+1} +\sqrt{z+2} + \sqrt{x+3} \\ \sqrt{y+1} + \sqrt{z+2} +\sqrt{x+3} = \sqrt{z+1}+\sqrt{x+2} + \sqrt{y+3} \end{matrix}
What is the value of $\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}$ ?
Thanks for helping solving this problem.
EDIT: I have done 
$(\sqrt{x+1} + \sqrt{y+2} +\sqrt{z+3})^2$=$ (\sqrt{y+1} +\sqrt{z+2} + \sqrt{x+3})^2$ and  $( \sqrt{y+1} + \sqrt{z+2} +\sqrt{x+3})^2 = (\sqrt{z+1}+\sqrt{x+2} + \sqrt{y+3})^2$ but haven't recived any helpful things.
 [/INST] The only answer that will satisfy the two equations involving square roots is that $$x=y=z$$. $$x$$,$$y$$,and$$z$$ can be any real numbers between -1 and 0 and above 0. If $$x=y=z$$, $$x-y=0$$, $$y-z=0$$, and $$z-x=0$$. So the result will be $$0/z+0/x+0/y$$ and since $$x,y, and z$$ do not equal zero. The resulting equation will equal zero $$0/z+0/x+0/y=0$$.
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