But Which Two Digits?

Since $a$ and $b$ are both single digit positive integers, let us list all 9 possibilities:

$1^3=1\\ 2^3=8\\ 3^3=27\\ 4^3=64\\ 5^3=125\\ 6^3=216\\ 7^3=343\\ 8^3=512\\ 9^3=729$

Notice that for $a=1,4,5,6,9$ , the last digit of $a^3$ is $a$ . We don't want this.

From the list above, we can see two possible pairs:

• $2^3$ ends with $8$ , and $8^3$ ends with $2$
• $3^3$ ends with $7$ , and $7^3$ ends with $3$

Therefore, $a=2,b=8;\;a=3,b=7$

Our answer is $2+8=3+7=\boxed{10}$

The conditions for $a$ and $b$ can be expressed as:
$\begin{aligned} a^3 &\equiv b \pmod{10} \\ b^3 &\equiv a \pmod{10} \end{aligned}$

After substituting every possible digit in place of $a$ in the first one, finding $b$ , checking it works for the second one, and eliminating the solution pairs in which $a = b$ , we notice that in every possible pair of solutions, $a + b = \boxed{10}$ which is the answer.