Question

Cho    \(\hept{\begin{cases}a^3-3ab^2=2\\b^3-3a^2b=-11\end{cases}}\)

Tính  \(a^2+b^2\)

Answer 1

\(\hept{\begin{cases}a^3-3ab^2=2\\b^3-3a^2b=-11\end{cases}\Rightarrow\hept{\begin{cases}\left(a^3-3ab^2\right)^2=4\\\left(b^3-3a^2b\right)^2=121\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}a^6-6a^4b^2+9a^2b^4=4\left(1\right)\\b^6-6a^2b^4+9a^4b^2=121\left(2\right)\end{cases}}\)

Cộng ( 1 ) với (2  ), ta được : \(a^6+b^6+3a^2b^4+3a^4b^2=125\)

\(\Rightarrow\left(a^2+b^2\right)^3=125\Rightarrow a^2+b^2=5\)