当x-y=1,求x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值
问题描述：
当x-y=1,求x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值
答
x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
=（x^4-x^3y）-xy^3+y^4-3x^2y+3xy^2
=x^3(x-y)-y^3(x-y)-3xy(x-y)
=x^3-y^3-3xy
=(x-y)(x^2+xy+y^2)-3xy
=x^2+xy+y^2-3xy
=x^2-2xy+y^2
=(x-y)^2
=1