Question

Cho \(^{a^3-3ab^2=2}\)

và \(b^3-3a^2b=-11\)

Tính \(a^2+b^2\)

Answer 1

Ta có: \(a^3-3ab^2=2\)

\(\Rightarrow\left(a^3-3ab^2\right)^2=4\)

\(\Leftrightarrow a^6-6a^4b^2+9a^2b^4=4\left(1\right)\)

Lại có: \(b^3-3a^2b=-11\)

\(\Rightarrow\left(b^3-3a^2b\right)=121\)

\(\Leftrightarrow b^6-6a^2b^4+9a^4b^2=121\left(2\right)\)

Lấy \(\left(1\right)+\left(2\right)\)ta được: 

\(a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)

\(\Leftrightarrow a^6+3a^4b^2+b^6+3a^2b^4=125\)

\(\Leftrightarrow\left(a^2+b^2\right)^3=125\)

\(\Leftrightarrow a^2+b^2=5\)

Vậy ...