Define $x\otimes y=x^3-y$. What is $h\otimes (h\otimes h)$?
$\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3$
By the definition of $\otimes$, we have $h\otimes h=h^{3}-h$. 
Then, $h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)=\boxed{\textbf{(C) }h}.$