Đặt \(\left\{{}\begin{matrix}a=\sqrt[3]{x+7}\\b=\sqrt[3]{2-x}\end{matrix}\right.\) \(\Rightarrow a^3+b^3=9\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=9\) (1)
Pt đã cho tương đương: \(a^2+b^2-ab=3\) (2)
Thay (2) vào (1) ta được:
\(3\left(a+b\right)=9\Rightarrow a+b=3\Rightarrow b=3-a\) (3)
Thay (3) vào (2) ta được:
\(a^2+\left(3-a\right)^2-a\left(3-a\right)-3=0\)
\(\Leftrightarrow a^2+a^2-6a+9-3a+a^2-3=0\) \(\Leftrightarrow3a^2-9a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x+7}=1\\\sqrt[3]{x+7}=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+7=1\\x+7=8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)
