Cho a^3 - 3ab^2 = 19 và b^3 -3a^2b= 98. Tính E = (a^2+b^2)^3
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\(a^3-3ab^2=19\Rightarrow\left(a^3-3ab^2\right)^2=361\)
\(\Leftrightarrow a^6-6a^4b^2+9a^2b^4=361\left(1\right)\)
\(b^3-3a^2b=98\Rightarrow\left(b^3-3a^2b\right)^2=9604\)
\(\Leftrightarrow b^6-6a^2b^4+9a^4b^2=9604\left(2\right)\)
\(\text{Công 2 vế (1) và (2) ta được :}\)
\(a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=9956\)
\(\Leftrightarrow a^6+3a^4b^2+3a^2b^4+b^6=9956\)
\(\Leftrightarrow\left(a^2+b^2\right)^3=9956\)
\(\Leftrightarrow a^2+b^2=\sqrt[3]{9956}\)
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bình phương 2 vế của cả 2 cái kia rồi trừ vế theo vế ..................
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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 !!!
Ta có : \(\left(a^3-3ab^2\right)^2=a^6-6a^4b^2+9a^2b^4.\)
Lại có : \(\left(b^3-3a^2b\right)^2=b^6-6a^2b^4+9a^4b^2\)
\(\Rightarrow\left(a^3-3ab^2\right)^2+\left(b^3-3a^2b\right)^2=19^2+98^2\)
\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=9965\)
\(\Rightarrow a^6+3a^4b^2+3a^2b^4+b^6=9965\)
\(\Rightarrow\left(a^2+b^2\right)^3=9965\)
\(\Rightarrow a^2+b^2=\sqrt[3]{9965}\)
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Ta có: \(\left\{{}\begin{matrix}a^3-3ab^2=19\\b^3-3a^2b=98\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\left(a^3-3ab^2\right)^2=19^2=361\\\left(b^3-3a^2b\right)^2=98^2=9604\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a^6-6a^4b^2+9a^2b^4=361\\b^6-6a^2b^4+9a^4b^2=9604\end{matrix}\right.\)
=> \(a^6+b^6+\left(9a^2b^4-6a^2b^4\right)+\left(9b^2a^4-6a^4b^2\right)=9965\)
=> \(a^6+3a^2b^4+3a^4b^2+b^6=9965\)
=> \(\left(a^2+b^2\right)^3=9965\)
=> \(a^2+b^2=\sqrt[3]{9965}\)