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E

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$P=a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)^2$

$a+b=a^2+b^2-ab \ge \dfrac{(a+b)^2}{2}-\dfrac{(a+b)^2}{4}=\dfrac{(a+b)^2}{4}$

\Leftrightarrow $0 \le a+b \le 4$

\Rightarrow $P \le 16$
 
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