Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$. What is $a-b$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$
Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$, the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$. Looking at the answer choices, the only multiple of $3$ is $\boxed{\textbf{(C)}\ 3}$.