Alternatively, let the numbers be $\displaystyle x$ and $\displaystyle y$. we have
$\displaystyle x^3 + y^3 = 5$
$\displaystyle x^2 + y^2 = 3$
So as Moo said, $\displaystyle 5 = (x + y)(3 - xy)$
that is, $\displaystyle \boxed{5 = 3(x + y) - (x^2y + xy^2)}$
Now, $\displaystyle (x + y)^3 = x^3 + y^3 + 3(x^2y + y^2x)$
which means $\displaystyle \boxed{5 = (x + y)^3 - 3(x^2y + xy^2)}$
Now, let $\displaystyle a = x+ y$ and $\displaystyle b = x^2y + xy^2$, then we have the system
$\displaystyle 3a - b = 5$ .....................(1)
$\displaystyle a^3 - 3b = 5$ ................(2)
Solve this system for $\displaystyle a$ and you have your answer (since $\displaystyle a$ is the sum of the two numbers)
I like your solution, Moo
But, of course, who said they had to be integers?