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Let x, y, z be real numbers each of whose absolute value is different from \displaystyle \frac{1}{\sqrt 3} such that x + y + z = xyz. Prove that
\frac{3x-x^{3}}{1-3x^{2}}+\frac{3y-y^{3}}{1-3y^{2}}+\frac{3z-z^{3}}{1-3z^{2}}=\frac{3x-x^{3}}{1-3x^{2}}\cdot\frac{3y-y^{3}}{1-3y^{2}}\cdot\frac{3z-z^{3}}{1-3z^{2}}

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