The expression $a^3-a^{-3}$ equals: 
$\textbf{(A) \ }\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right) \qquad \textbf{(B) \ }\left(\frac{1}{a}-a\right)\left(a^2-1+\frac{1}{a^2}\right) \qquad \textbf{(C) \ }\left(a-\frac{1}{a}\right)\left(a^2-2+\frac{1}{a^2}\right) \qquad$
$\textbf{(D) \ }\left(\frac{1}{a}-a\right)\left(\frac{1}{a^2}+1+a^2\right) \qquad \textbf{(E) \ }\text{none of these}$
Recall that $x^3-y^3=(x-y)(x^2+xy+y^2)$. Letting $a=x$ and $a^{-1}=y$, we find that $a^3-a^{-3}= \boxed{\textbf{(A)}\ \left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right)}$.