Đặt \(\left\{{}\begin{matrix}\sqrt[3]{6+x}=a\\\sqrt[3]{3-x}=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+b+ab=5\\a^3+b^3=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=5\\\left(a+b\right)^3-3ab\left(a+b\right)=9\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a+b=u\\ab=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^3-3uv=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}v=5-u\\u^3-3uv=9\end{matrix}\right.\)
\(\Rightarrow u^3-3u\left(5-u\right)=9\)
\(\Leftrightarrow u^3+3u^2-15u-9=0\)
\(\Leftrightarrow\left(u-3\right)\left(u^2+6u+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=2\\u=-3+\sqrt{6}\Rightarrow v=8-\sqrt{6}\left(l\right)\\u=-3-\sqrt{6}\Rightarrow v=8+\sqrt{6}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\ab=2\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(1;2\right);\left(2;1\right)\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{6+x}=1\\\sqrt[3]{6+x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
