已知x的平方-5x-1997=0,求代数式（x-2）的3次方-（x-1）的平方+1 除以x-2的值.

(x-2)(x-2)^2-(x-1)^2+1/(x-2)
=(x-2)(x-2)^2-[(x-1)^2-1]/(x-2)
=(x-2)(x-2)^2-[(x-1+1)*(x-1-1)]/(x-2)
=(x-2)(x-2)^2-[x*(x-2)]/(x-2)
=(x-2)[(x-2)^2-x]/(x-2)
=(x-2)^2-x
=x^2-5x+4
因为x^2-5x-1997=0
所以x^2-5x=1997
所以x^2-5x+4=1997+4=2001