\(\hept{\begin{cases}x^2+xy+y^2=3\left(1\right)\\x^3+3\left(y-x\right)=1\end{cases}}\)
\(\Rightarrow x^3+\left(y-x\right)\left(x^2+xy+y^2\right)=1\)
\(\Leftrightarrow x^3+y^3-x^3=1\)
\(\Leftrightarrow y^3=1\)
\(\Leftrightarrow y=1\)
Thế vô pt (1) được \(x^2+x+1=3\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)