IMO 1974 SL 9

Problem

II 3 (CUB 3) Let x, y, z be real numbers each of whose absolute value is different from 1/ \sqrt 3 such that x + y + z = xyz. Prove that 3x −x3 1 −3x2 + 3y −y3 1 −3y2 + 3z −z3 1 −3z2 = 3x −x3 1 −3x2 \cdot 3y −y3 1 −3y2 \cdot 3z −z3 1 −3z2 .

Solution

There exist real numbers a, b, c with tan a = x, tan b = y, tan c = z. Then using the additive formula for tangents we obtain tan(a + b + c) = x + y + z −xyz 1 −xy −xz −yz . We are given that xyz = x + y + z. In this case xy + yz + zx = 1 is impossible; otherwise, x, y, z would be the zeros of a cubic polynomial t3 −\lambdat2 + t −\lambda = (t2 + 1)(t −\lambda) (where \lambda = xyz), which has only one real root. It follows that x + y + z = xyz ⇐⇒ tan(a + b + c) = 0. (1) Hence a + b + c = k\pi for some k \inZ. We note that 3x−x3 1−3x2 actually expresses tan 3a. Since 3a + 3b + 3c = 3k\pi, the result follows from (1) for the numbers 3x−x3 1−3x2 , 3y−y3 1−3y2 , 3z−z3 1−3z2 .