AIME 2015 Problem 8

Because we know that $a,b$ are positive integers, we know that both the numerator and denominator of our fraction $\frac{ab+1}{a+b}$ are positive, so we can legitimately rearrange by cross multiplying. We should get that

$2ab+2<3a+3b \\ 4ab+4-6a-6b<0 \\ (2a-3)(2b-3)<5$

We can check for all possible combinations for positive integers $a,b$ that satisfy this inequality. We should get all ordered pairs $(a,b)$ are

$(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2)$

Since $a$ and $b$ are interchangeable, we can reduce our cases to

$(1,1),(2,1),(3,1),(2,2),(2,3)$

We can also see that if $a$ or $b$ is 1, then the expression $\frac{a^{3}b^{3}+1}{a^{3}+b^{3}}$ equals 1. This reduces our cases even further to

$(2,3),(2,2)$

Checking $(2,2)$ , $\frac{a^{3}b^{3}+1}{a^{3}+b^{3}}=\frac{65}{16}$

Checking $(2,3)$ , $\frac{a^{3}b^{3}+1}{a^{3}+b^{3}}=\frac{217}{35}=\frac{31}{5}$

Since $\frac{31}{5}>\frac{65}{16}>1$ , our answer is $31+5=\boxed{36}$