若x+y=-1,则x^4+5yx^3+yx^2+8x^2*y^2+xy^2+5xy^3+y^4的值等于多少

题目

若x+y=-1,则x^4+5yx^3+yx^2+8x^2*y^2+xy^2+5xy^3+y^4的值等于多少

答案

x^4+5yx^3+yx^2+8x^2*y^2+xy^2+5xy^3+y^4
=(x^4+y^4+8x^2*y^2)+5xy(x^2+y^2)+xy(x+y)
=(x^2+y^2)^2+6x^2*y^2+5xy(x^2+y^2)+xy(x+y)
=(x^2+y^2)^2+4x^2*y^2+4xy(x^2+y^2)+2x^2*y^2+xy(x^2+y^2)+xy(x+y)
=[(x^2+y^2)+2xy]^2+xy(2xy+x^2+y^2+x+y)
=[(x+y)^2]^2+xy[(x+y)^2+(x+y)]
=(x+y)^4+xy[(x+y)^2+(x+y)]
由于x+y=-1
所以原式=(-1)^4+xy[(-1)^2+(-1)]=1+0=1