jmc

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 $\boxed{\frac{31}{5}}.$