Thêm bớt hạng tử thôi:
\(x^3+3xy+y^3-1\)
\(=\left(x^3+3x^2y+3xy^2+y^3-1\right)-3x^2y-3xy^2+3xy\)
\(=\left[\left(x+y\right)^3-1^3\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)