$$\frac{a}{b}+\frac{b}{a}-a b=\frac{a^{2}+b^{2}}{a b}-(a-b)=\frac{a^{2}+b^{2}}{a-b}-\frac{(a-b)^{2}}{(a-b)}=\frac{2 a}{a-b}=\frac{2(a-b)}{a-b}=2 \Rightarrow \text{E}.$$ 
Two non-zero real numbers, $a$ and $b$, satisfy $a b=a-b$. Which of the following is a possible value of $\frac{a}{b}+\frac{b}{a}-a b$? (2000 AMC 12 Problems, Question \#11) 

A) $-2$ 
B) $\frac{-1}{2}$ 
C) $\frac{1}{3}$ 
D) $\frac{1}{2}$ 
E) $$2$$ 