Ta có :
x3-x+y3-y
=(x3+y3)-(x+y)
=(x+y)(x2-xy+y2)-(x+y)
=(x+y)(x2-xy+y2-1)
Ta có :
\(x^3-x+y^3-y=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\cdot\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\cdot\left(x^2-xy+y^2-1\right)\)
Vậy \(x^3-x+y^3-y=\left(x+y\right)\cdot\left(x^2-xy+y^2-1\right).\)