Find all complex numbers x, y, z which satisfy x + y + z = x2 + y2 + z2 = x3 + y3 + z3 = 3.
Solution
Answer: 1, 1, 1.
We have (x + y + z)2 = (x2 + y2 + z2) + 2(xy + yz + zx), so xy + yz + zx = 3.
(x + y + z)3 = (x3 + y3 + z3) + 3(x2y + xy2 + y2z + yz2 + z2x + zx2) + 6xyz, so 8 = (x2y + xy2 + y2z + yz2 + z2x + zx2) + 2xyz. But (x + y + z)(xy + yz + zx) = (x2y + xy2 + y2z + yz2 + z2x + zx2) + 3xyz, so xyz = -1. Hence x, y, z are the roots of the cubic w3 - 3w2 + 3w - 1 = (w - 1)3. Hence x = y = z = 1.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002