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Prove the identity if x+y+z=xyz

Source:

9/22/2010
Let x,y,zx, y, z be real numbers each of whose absolute value is different from 13\frac{1}{\sqrt 3} such that x+y+z=xyzx + y + z = xyz. Prove that 3x−x31−3x2+3y−y31−3y2+3z−z31−3z2=3x−x31−3x2⋅3y−y31−3y2⋅3z−z31−3z2\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}
trigonometryalgebraTrigonometric Identitiestrigonometric substitutionequationIMO Shortlist