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1987 USAMO Problems/Problem 1

Revision as of 10:17, 20 December 2015 by Bassmaster (talk | contribs)

By expanding you get \[m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3\] From this the two $m^3$ cancel and you get: \[2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0\] You can divide by $n$ (nonzero). You get: \[2n^2+m+m^2n+3m^2-3mn=0\] You can now factor the equation into: \[2n^2+(m^2-3m)n+(3m^2+m)=0\]

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