设：x^3-1=(x-1)(ax^2+bx+1)
(x-1)(ax^2+bx+1)
=ax^3+bx^2+x-ax^2-bx-1
=ax^3+(b-a)x^2+(1-b)x-1
对比系数可得:
b-a=0
1-b=0
所以,a=b=1
故,x^3-1=(x-1)(x^2+x+1)