当x-y=1,求x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值

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当x-y=1,求x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值

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