Question 9
II 3 (CUB 3)
Let $x,y,z$ be real numbers each of whose absolute value is different from $1/\sqrt3$ such that $x+y+z=xyz$. Prove that
$$\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}\cdot\frac{3y-y^3}{1-3y^2}\cdot\frac{3z-z^3}{1-3z^2}.$$
Let $x,y,z$ be real numbers each of whose absolute value is different from $1/\sqrt3$ such that $x+y+z=xyz$. Prove that
$$\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}\cdot\frac{3y-y^3}{1-3y^2}\cdot\frac{3z-z^3}{1-3z^2}.$$