Suppose that $a$ and $b$ are positive integers such that $a-b=6$ and $\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9$. Find the smallest possible value of $b$.
Recall that $a^3+b^3 = (a+b)(a^2-ab+b^2)$. By the Euclidean Algorithm, we obtain: \begin{align*}
\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) &= \text{gcd}(a^2-ab+b^2, ab) \\
&= \text{gcd}(a^2-2ab+b^2, ab) \\
&= \text{gcd}((a-b)^2, ab) \\
&= \text{gcd}(36, ab).
\end{align*}Thus, $\text{gcd}(36, ab) = 9$. Trying values of $b$,  we find that $b = 1 \Rightarrow a=7$ and $ab = 7\Rightarrow \text{gcd}(36, ab) = 1$. If $b = 2$, then $a=8$ and $ab=16 \Rightarrow \text{gcd}(36, ab) = 4$. Finally, $b = 3 \Rightarrow a=9$ and $ab=27 \Rightarrow \text{gcd}(36, ab) = 9$. Therefore, the minimum possible value for $b$ is  $\boxed{3}$.