若代数式x^3+y^3+3x^2y+axy^2 含有因式x-y ，则 a=?，在实数范围内将这个代数式分解因式=?

x^3+y^3+3x^2y+axy^2=（x-y)(x^2+bxy-y^2）
（x-y)(x^2+bxy-y^2）
=x^3+bx^2y-xy^2-x^2y-bxy^2+y^3
=x^3+y^3+（b-1）x^2y-（b+1）xy^2
b-1=3，且-（b+1）=a
则b=4，a=-5
所以x^3+y^3+3x^2y+axy^2
=x^3+y^3+3x^2y-5xy^2
=（x-y)(x^2+4xy-y^2）
=（x-y)(x^2+4xy+4y^2-5y^2）
=（x-y)[(x+2y)^2-(√5y)^2]
=（x-y)(x+2y-√5y)(x+2y+√5y)
=（x-y)[x+(2-√5)y][x+(2+√5)y]

x^3+y^3+3x^2y+axy^2 含有因式x-y
则当x=y时,x^3+y^3+3x^2y+axy^2=0
5x^3+ax^3=0
a=-5
x^3+y^3+3x^2y-5xy^2
=(x-y)(x^2+4xy-y^2)
=(x-y)[x+(2+√5)y][x-(√5-2)y]

题型不错，可采用待定系数法分解因式

因为代数式x^3+y^3+3x^2y+axy^2 含有因式x-y
所以设x^3+y^3+3x^2y+axy^2=（x-y)(x^2+bxy-y^2）
而（x-y)(x^2+bxy-y^2）
=x^3+bx^2y-xy^2-x^2y-bxy^2+y^3
=x^3+y^3+（b-1）x^2y-（b+1）xy^2
所以令对应系数相等：b-1=3，且-（b+1）=a
则b=4，a=-5
所以x^3+y^3+3x^2y+axy^2
=x^3+y^3+3x^2y-5xy^2
=（x-y)(x^2+4xy-y^2）
=（x-y)(x^2+4xy+4y^2-5y^2）
=（x-y)[(x+2y)^2-(√5y)^2]
=（x-y)(x+2y-√5y)(x+2y+√5y)