If (x+\frac{1}{yz})-(y+\frac{1}{zx})=(y+\frac{1}{zx})-(z+\frac{1}{xy}) and x+z \neq 2y, then what is xyz equal to?

Explanation

The given equality implies the terms are in arithmetic progression. Difference d = (x-y) + \frac{x-y}{xyz} = (y-z) + \frac{y-z}{xyz}. Factoring gives (x-y)(1 + \frac{1}{xyz}) = (y-z)(1 + \frac{1}{xyz}) \implies (x-2y+z)(1 + \frac{1}{xyz}) = 0. Since x+z \neq 2y, we get 1 + \frac{1}{xyz} = 0 \implies xyz = -1.

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