\(\Leftrightarrow x^3+3x^2+3x+1-y^3+3\left(x+1-y\right)=0\)
\(\Leftrightarrow\left(x+1\right)^3-y^3+3\left(x+1-y\right)=0\)
\(\Leftrightarrow\left(x+1-y\right)\left(\left(x+1\right)^2+y\left(x+1\right)+y^2+3\right)=0\)
\(\Leftrightarrow x+1=y\)
(do \(\left(x+1\right)^2+y\left(x+1\right)+\frac{y^2}{4}+\frac{3y^2}{4}+3\)
\(=\left(x+1+\frac{y}{2}\right)^2+\frac{3y^2}{4}+3>0\) nên ngoặc sau luôn dương)
Thay vào pt dưới:
\(x^2+\left(x+1\right)^2-3x=1\)