ILL 1974 Problem 8

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