The inequality $ \frac{ab + 1}{a + b} < \frac{3}{2} $ turn into \[ab + 1 < \frac{3}{2} a + \frac{3}{2} b.\]Then \[ab - \frac{3}{2} a - \frac{3}{2} b + 1 < 0.\]Applying Simon's Favorite Factoring Trick, we get \[\left( a - \frac{3}{2} \right) \left( b - \frac{3}{2} \right) < \frac{5}{4}.\]Hence, \[(2a - 3)(2b - 3) < 5.\]If $ a = 1, $ then the inequality becomes \[3 - 2b < 5,\]which is satisfied for any positive integer $ b $. Similarly, if $ b = 1, $ then the inequality is satisfied for any positive integer $ a $. Otherwise, $ a \ge 2 $ and $ b \ge 2, $ so $ 2a - 3 \ge 1 $ and $ 2b - 3 \ge 1 $. Note that both $ 2a - 3 $ and $ 2b - 3 $ are odd, so $ (2a - 3)(2b - 3) $ is odd, so their product can only be 1 or 3. This leads us to the solutions $ (a,b) = (2,2), $ $ (2,3), $ and $ (3,2) $. If $ a = 1, $ then \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{b^3 + 1}{1 + b^3} = 1.\]Similarly, if $ b = 1, $ then the expression also simplifies to 1. For $ (a,b) = (2,2), $ \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{2^3 \cdot 2^3 + 1}{2^3 + 2^3} = \frac{65}{16}.\]For $ (a,b) = (2,3) $ or $ (3,2), $ \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{2^3 \cdot 3^3 + 1}{2^3 + 3^3} = \frac{31}{5}.\]Hence, the largest possible value of the expression is $ \boxed{\frac{31}{5}} $.