Factorials and cubes

Clearly $a=5,b=5$ is a solution. Now, to see if there are other solutions, we first note that the solutions for $a$ must be $>5$ . Then the unit's digit of $a!+5$ must be $5$ which implies that the unit's digit of $b$ also must be $5$ . Now, the $10$ 's digit of any number of the form $x^3$ where $x$ is an integer with $5$ as its unit's digit, is $\in \{2,7\}$ . Since the $10$ 's digit of $x!$ for any $x\ge 10$ is always $0$ , this implies that the other possible solutions for $a$ must $\in \{6,7,8,9\}$ . Now an easy check reveals that there is no other solution for $a$ and hence the only possible solution is $\boxed{a=5,b=5}$ .