AlgebraDifficulty 4.9Prove itNMO Selection Tests for JBMO · Romania
Let x and y be real nonzero numbers, such that x3+y3+3x2y2=x3y3. Determine the set of the possible values of E=x1+y1.
This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.
Rewrite the given condition successively (x+y)3−3xy(x+y)=x3y3−3x2y2, i.e. (x+y)3−(xy)3=3xy(x+y)−3x2y2, or (x+y−xy)(x2+2xy+y2+x2y+xy2+x2y2)=3xy(x+y−xy). We either have x+y=xy, which leads to E=1 (obtained for x=y=2), or x2+2xy+y2+x2y+xy2+x2y2=3xy. The last equality, multiplied by 2, can be written equivalently x2(y+1)2+y2(x+1)2+(x−y)2=0, which is possible if and only if x=y=−1 (x=y=0 is forbidden). In this case, E=−2. In conclusion, the set of the possible values of E is {−2,1}.
Source: MathNet,
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