x3-y3=(x-y)(x2+xy+y2)x2-y2=(x-y)(x+y)因此，原式=（x-y）（x2+xy+y2+x+y）
解：原式=x³-y³+x²-y²=(x-y)(x²+xy+y²)+(x+y)(x-y)=(x-y)(x²+xy+y²+x+y)
x3+x2-y3-y2
=x3-y3+x2-y2
=(x-y)(x2+xy+y2)+(x-y)(x+y)
=(x-y)(x+y+x2+xy+y2)
