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mathisfun1 · #2 August 13th, 2007, 06:18 AM

All permutations of $(k, -k, j)$ for integers k and j are solutions. To see that these are the only solutions, rewrite as $y^3+z^3 = (x+y+z)^3 - x^3$ which, assuming $y \ne -z$ , reduces to $x^2+(y+z)x+yz = 0 \Rightarrow x = -y$ or $x = -z$ with the other variable arbitrary, giving the solutions $(k, -k, j), \ (k, j, -k)$ and, by symmetry in the original equation, $(j, k, -k)$