Question

[Nguyễn Thúy linh]

cho \(a^3-3ab^2=19\)

\(b^3-3a^2b=98\)

Tính \(P=a^2+b^2\)

Answer 1

[Kudo Shinichi]

Ta có : \(\hept{\begin{cases}a^3-3ab^2=19\\b^3-3a^2b=98\end{cases}\Rightarrow\hept{\begin{cases}\left(a^3-3ab^2\right)^2=19^2=361\\\left(b^3-3a^2b\right)^2=98^2=9604\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}a^6-6a^4b^2+9a^2b^4=361\\b^6-6a^2b^4+9a^4b^2=9604\end{cases}}\)

\(\Rightarrow a^6+b^6+\left(9a^2b^4-6a^2b^4\right)+\left(9b^2a^4-6a^4b^2\right)=9965\)

\(\Rightarrow a^6+3a^2b^4+3a^4b^2+b^6=9965\)

\(\Rightarrow\left(a^2+b^2\right)^3=9965\)

\(\Rightarrow a^2+b^2=\sqrt[3]{9965}\)

Chúc bạn học tốt !!!

Answer 2

[Phạm Thị Thùy Linh]

Ta có :  \(\left(a^3-3ab^2\right)^2=a^6-6a^4b^2+9a^2b^4.\)

Lại có : \(\left(b^3-3a^2b\right)^2=b^6-6a^2b^4+9a^4b^2\)

\(\Rightarrow\left(a^3-3ab^2\right)^2+\left(b^3-3a^2b\right)^2=19^2+98^2\)

\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=9965\)

\(\Rightarrow a^6+3a^4b^2+3a^2b^4+b^6=9965\)

\(\Rightarrow\left(a^2+b^2\right)^3=9965\)

\(\Rightarrow a^2+b^2=\sqrt[3]{9965}\)