Đề đúng (Hậu Giang 2013-2014) :Cho \(a^3+3ab^2=2014\)và \(b^3+3a^2b=2013\).Tính \(P=a^2-b^2\)
Ta có:
\(\left(a+b\right)^3=a^3+b^3+3a^2b+3ab^2=\left(a^3+3ab^2\right)+\left(b^3+3a^2b\right)=2014+2013=4027\)
\(\Rightarrow a+b=\sqrt[3]{4027}\)
\(\left(a-b\right)^3=a^3+3ab^2-\left(b^3+3a^2b\right)=2014-2013=1\)
\(\Rightarrow a-b=1\)
do đó \(P=a^2-b^2=\left(a+b\right)\left(a-b\right)=1.\sqrt[3]{4027}=\sqrt[3]{4027}\)