jmc

Recall that $a^3+b^3=(a+b)(a^2-ab+b^2)$. By the Euclidean Algorithm, we obtain: $$\begin{array}{rl}\text{gcd}\left(\frac{a^3+b^3}{a+b},ab\right)& =\text{gcd}(a^2-ab+b^2,ab)\\ & =\text{gcd}(a^2-2ab+b^2,ab)\\ & =\text{gcd}((a-b{)}^2,ab)\\ & =\text{gcd}(36,ab).\end{array}$$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 $\boxed{3}$.