若代数式x 3 +y 3 +3x 2 y+axy 2 含有因式x-y,则a=___,在实数范围内将...
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∵代数式x 3 +y 3 +3x 2 y+axy 2 含有因式x-y,
∴当x=y时,x 3 +y 3 +3x 2 y+axy 2 =0,
∴令x=y,即x 3 +x 3 +3x 3 +ax 3 =0,
则有5+a=0,解得a=-5.
将a=-5代入x 3 +y 3 +3x 2 y+axy 2 ,得
x 3 +y 3 +3x 2 y-5xy 2
=x 3 -x 2 y+4x 2 y-5xy 2 +y 3
=(x-y)x 2 +y(x-y)(4x-y)
=(x-y)(x 2 +4xy-y 2 )
= (x-y)(x+2y+ 5 y)(x+2y- 5 y) .
故答案为: (x-y)(x+2y+ 5 y)(x+2y- 5 y) .