\(2x^3-2y^3+5xy+1=0\\\leftrightarrow 2(x-y)(x^2+xy+y^2)+5xy+1=0\\\leftrightarrow 2(x-y) [(x-y)^2 +3xy] +5xy +1=0(*)\)

Đặt \(a=x-y,b=xy (a,b\in Z)\)

\((*)\leftrightarrow 2a (a^2+3b) +5b+1=0\\\leftrightarrow 2a^3+6ab +5b+1=0\\\leftrightarrow 2a^3+1+b(6a+5)=0\\\leftrightarrow 2a^3+1=-b(6a+5)\\\leftrightarrow -b=\dfrac{2a^3+1}{6a+5}\left(a\ne \dfrac{-5}{6}\right)\\-b\in Z\to \dfrac{2a^3+1}{6a+5}\in Z\\\to (2a^3+1)\vdots (6a+5)\\\to 108 (2a^3+1)\vdots (6a+5)\\\to (216a^3+108)\vdots (6a+5)\\\to [(216a^3+125) -17]\vdots (6a+5)\\\to 17\vdots (6a+5)\\\to 6a+5=-1,6a+5=1,6a+5=17, 6a+5=-17\\\to a=-1, a=\dfrac{-4}{6}\text{(L)},a=2,a=\dfrac{-22}{6}\text{(L)}\\\bullet a=-1\\\to -b=1\to b=-1\\\bullet a=2\to -b=1\to b=-1\\\to (a;b)=(-1;-1), (2;-1)\\\bullet (a;b)=(-1;-1)\\\to x-y=-1,xy=-1\\\to x=y-1, (y-1)y=-1\\(y-1)y=-1\\\to y^2-y+1=0\text{(VN)}\\\bullet (a;b)=(2;-1)\\\to x-y=2,xy=-1\\\to x=y+2,(y+2)y=-1\\(y+2)y=-1\\\to y^2+2y+1=0\\\to y=-1\to x=1\text{(Tm)}\)

Thử lại cặp \((x;y)=(1;-1)\) ta thấy thỏa mãn.

Pt có nghiệm \((x;y)=(1;-1)\)