当x=2,y=1时,代数式x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
问题描述：
当x=2,y=1时,代数式x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
答
当x=2,y=1时，
代数式x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
=2^4-2*1^3-2^3*1-3*2^2*1+3*2*1^2+1^4
=16-2-8-12+6+1
=1
答
x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
=x^4-x^3y-3xy(x-y)+y^4-xy^3
=x^3(x-y)-3xy(x-y)-y^3(x-y)
=(x-y)(x^3-3xy-y^3)
=(2-1)(8-6-1)
=1*1
=1