已知x^2-3x-2=0
即：x^2-3x-2=(x^2-2x+1)-x-3=(x-1)^2-(x-1)-4=0
所以有：(x-1)^2-(x-1)=4……①
又分子：
(x-1)^3-x^2+1
=(x-1)^3-x^2+2x-1-2x+1+1
=(x-1)^3-(x^2-2x+1)-2x+2
=(x-1)^3-(x-1)^2-2(x-1)
=(x-1)[(x-1)^2-(x-1)]-2(x-1)
[将(x-1)^2-(x-1)=4……①代入]
=4(x-1)-2(x-1)
=2(x-1)
所以原式：
[(x-1)^3-x^2+1]/(x-1)
=2(x-1)/(x-1)
=2
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