题目

已知x-y=1 求代数式x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值

提问时间：2020-07-21

答案

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)而x-y=1=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