Đặt \(\left(\sqrt[3]{x};\sqrt[3]{y}\right)\rightarrow\left(a;b\right)\) thì :
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\sqrt[3]{x}+\sqrt[3]{y}=a+b=1\\x-y=a^3-b^3=9\end{matrix}\right.\)
Xét \(a+b=1\Rightarrow a=1-b\)
Xét \(a^3-b^3=9\Rightarrow\left(a-b\right)\left(a^2+ab+b^2\right)=9\left(1\right)\)
Thay \(a=1-b\) vào \(\left(1\right)\) có:
\(\left(1\right)\Leftrightarrow-2b^3+3b^2-3b-8=0\)
\(\Leftrightarrow-\left(b+1\right)\left(2b^2-5b+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b+1=0\\2b^2-5b+8=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}b=-1\\2\left(b-\dfrac{5}{4}\right)^2+\dfrac{39}{8}>0\forall b\end{matrix}\right.\)
Thay \(b=-1\) vào \(a=1-b\) có:
\(a=1-b=1-(-1)=2\)
Vậy hpt có nghiệm là \(\left(a;b\right)=\left(2;-1\right)\)
