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Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt[3]{x}}=a\\\frac{1}{\sqrt[3]{y}}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=9\\\left(a+b\right)\left(a+1\right)\left(b+1\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b\right)=9\\\left(a+b\right)\left(ab+a+b+1\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b\right)=9\\ab\left(a+b\right)+\left(a+b\right)^2+\left(a+b\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b\right)=9\\3ab\left(a+b\right)+3\left(a+b\right)^2+3\left(a+b\right)=54\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^3+3\left(a+b\right)^2+3\left(a+b\right)=63\)
\(\Leftrightarrow\left(a+b\right)^3+3\left(a+b\right)^2+3\left(a+b\right)+1=64\)
\(\Leftrightarrow\left(a+b+1\right)^3=4^3\)
\(\Leftrightarrow a+b+1=4\Rightarrow a+b=3\)
\(\Rightarrow3\left(ab+3+1\right)=18\Rightarrow ab=2\)
Theo Viet đảo; a và b là nghiệm:
\(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(1;2\right);\left(2;1\right)\Rightarrow\left(x;y\right)=\left(1;\frac{1}{8}\right);\left(\frac{1}{8};1\right)\)
