#Yup it's number theory

a=1; b=1;c=gcd((a^3+b^3)/(a+b), (a*b)); if(c==9 and a-b==6, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;

OUTPUT:  a = 9   and   b = 3

I'm not too savvy with number theory, but maybe you could try using the euclidean algorithm with this problem?

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.

\(a-b=6\\ a=6+b\)

\(\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\\ \text{gcd}\left(\frac{(a+b)(a^2-ab+b^2)}{a+b}, ab\right) = 9\\ \text{gcd}\left((a^2-ab+b^2), ab\right) = 9\\ \text{gcd}\left((a^2-2ab+b^2+ab), ab\right) = 9\\ \text{gcd}\left(((a-b)^2+ab), ab\right) = 9\\ \text{gcd}\left((6^2+ab), ab\right) = 9\\ \text{gcd}\left((36+ab), ab\right) = 9\\ \text{gcd}\left(9(4+\frac{ab}{9}), ab\right) = 9\\ \text{gcd}\left((4+\frac{ab}{9}), \frac{ab}{9}\right) = 1\\\)

So ab is a multiple of 9 AND ab is not even  AND   \(\frac{ab}{9}\)  is not even

If ab=9 then gcd is correct, but there are not values of a and b exactly 6 units apart

If ab=27 then gcd is correct, and 9*3=27,   9-3=6

So the smallest values of a and b are   9 and 3 respectfully. 

The smallest value of b is 3.

LaTex:

\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\\
\text{gcd}\left(\frac{(a+b)(a^2-ab+b^2)}{a+b}, ab\right) = 9\\
\text{gcd}\left((a^2-ab+b^2), ab\right) = 9\\
\text{gcd}\left((a^2-2ab+b^2+ab), ab\right) = 9\\
\text{gcd}\left(((a-b)^2+ab), ab\right) = 9\\
\text{gcd}\left((6^2+ab), ab\right) = 9\\
\text{gcd}\left((36+ab), ab\right) = 9\\
\text{gcd}\left(9(4+\frac{ab}{9}), ab\right) = 9\\
\text{gcd}\left((4+\frac{ab}{9}), \frac{ab}{9}\right) = 1\\