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

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

原题应该是：x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
由题可得：x-y=1,y-x=-1
原式= x^4-x^3y+ y^4- xy^3-3x^2y+3xy^2
=x^3(x-y)+y^3(y-x)-3xy(x-y)
=x^3-y^3-3xy
=(x-y)(x^2+xy+y^2)-3xy
=(x^2+xy+y^2)-3xy
=(x-y)^2
=1^2=1

原题应该是：x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
由题可得：x-y=1,y-x=-1
原式= x^4-x^3y+ y^4- xy^3-3x^2y+3xy^2
=x^3(x-y)+y^3(y-x)-3xy(x-y)
=x^3-y^3-3xy
=(x-y)(x^2+xy+y^2)-3xy
=(x^2+xy+y^2)-3xy
=(x-y)^2
=1^2=1