Cộng vế với vế:
\(x^3+3x^2y+3xy^2+y^3=4039\)
\(\Leftrightarrow\left(x+y\right)^3=4039\Rightarrow x+y=\sqrt[3]{4039}\)
Trừ vế cho vế:
\(x^3-3x^2y+3xy^2-y^3=1\)
\(\Leftrightarrow\left(x-y\right)^3=1\Rightarrow x-y=1\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=\sqrt[3]{4039}\\x-y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{\sqrt[3]{4039}+1}{2}\\y=\frac{\sqrt[3]{4039}-1}{2}\end{matrix}\right.\)
\(\Rightarrow x^2+y^2=...\)
Tính \(x^2-y^2\) thì kết quả đẹp hơn