Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+11}=a\\\sqrt[3]{x+2}=b\end{matrix}\right.\) . Ta có hệ phương trình :
\(\left\{{}\begin{matrix}a-b=3\\a^3-b^3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\\left(b+3\right)^3-b^3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\9b^2+27b+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\\left(b+1\right)\left(b+2\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\\\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\end{matrix}\right.\)
Với \(a=2;b=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{x+11}=2\\\sqrt[3]{x+2}=-1\end{matrix}\right.\Rightarrow x=-3\)
Với \(a=1;b=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{x+11}=1\\\sqrt[3]{x+2}=-2\end{matrix}\right.\Rightarrow x=-10\)
Vậy \(S=\left\{-10;-3\right\}\)
