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Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Numbers aa, bb fulfill both equalities simultaneously:
a2+b2=1 and a3+b3=1. a^2 + b^2 = 1 \text{ and } a^3 + b^3 = -1.

What is the possible value of the expression a3+b2a^3 + b^2?

Solution

From the first equation 1a1-1 \leq a \leq 1 and 1b1-1 \leq b \leq 1, therefore 01+a20 \leq 1+a \leq 2 and 01+b20 \leq 1+b \leq 2. Add both equations and get
a2(1+a)+b2(1+b)=0. a^2(1+a) + b^2(1+b) = 0.
As both items are non-negative, their sum equals zero if and only if every item equals 00. So a,b{1,0}a, b \in \{-1, 0\}. From the first equation it follows that one variable is not zero, from the second equation it follows that this variable equals 1-1. Thus, the equations are satisfied by such pairs of numbers (a,b):(1,0)(a, b): (-1, 0) and (0,1)(0, -1), hence a3+b2=1a^3 + b^2 = -1 or a3+b2=1a^3 + b^2 = 1.

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