x3-x+3x2y+3xy2+y3-y
=x2(x-1)+3(x2y+xy2)+y2(y-1)
=x2(x-1)+3(x2.y+y2.x)+y2(y-1)
=x2(x-1)+3{[x(x+1)+y(y+1)]}+y2(y-1)
=x2(x-1)+3.x(x+1)+3.y(y+1)+y2(y-1)
=x2(x-1)+2x2+3.x(x+1)+3.y(y+1)+y2(y-1)+2y2-2x2-2y2
=x2(x+1)+3.x(x+1)+3.y(y+1)+y2(y+1)-2x2-2y2
=(x2+3)(x+1)+(y2+3)(y+1)-2(x2+y2)
ta có : (x*3+3x*2y+3xy*2+y*3)-(x+y)
=(x+y)*3-(x+y)
=(x+y)((X+Y)*2-1)
(x+y)(x+y+1)(x+Y-1)
\(x^3-x+3x^2y+3xy^2+y^3-y=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\cdot\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\cdot\left(x+y+1\right)\cdot\left(x+y-1\right)\)
