jmc

Notice that by the sum of cubes factorization, $n^3+8=(n+2)(n^2-2n+4)$ is an integer divisible by $n+2$. Thus, $$\begin{array}{rl}\text{gcd}(n^3+9,n+2)& =\text{gcd}(n^3+9-(n^3+8),n+2)\\ & =\text{gcd}(1,n+2)\\ & =\boxed{1}.\end{array}$$