若代数式x3+y3+3x2y+axy2含有因式x-y,则a=______,在实数范围内将这个代数式分解因式,得x3+y3+3x2y+axy2=______.-数学

题目简介

若代数式x3+y3+3x2y+axy2含有因式x-y,则a=______,在实数范围内将这个代数式分解因式,得x3+y3+3x2y+axy2=______.-数学

题目详情

若代数式x3+y3+3x2y+axy2含有因式x-y,则a=______,在实数范围内将这个代数式分解因式,得x3+y3+3x2y+axy2=______.
题型:填空题难度:中档来源:不详

答案

∵代数式x3+y3+3x2y+axy2含有因式x-y,
∴当x=y时,x3+y3+3x2y+axy2=0,
∴令x=y,即x3+x3+3x3+ax3=0,
则有5+a=0,解得a=-5.
将a=-5代入x3+y3+3x2y+axy2,得
x3+y3+3x2y-5xy2
=x3-x2y+4x2y-5xy2+y3
=(x-y)x2+y(x-y)(4x-y)
=(x-y)(x2+4xy-y2)
=(x-y)(x+2y+
5
y)(x+2y-
5
y)

故答案为:(x-y)(x+2y+
5
y)(x+2y-
5
y)

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