Finding all integer solutions to $A\times B\times C = A! + B!+ C!$ with $0\leq A, B, C\leq 9$ without trying every combination?

We may suppose that $A\le B\le C$. Then $ABC\le C^3$. But if $C\ge 6$, then $C!>C^3$, so $$ABC<A!+B!+C!$$ Hence $C\le 5$. And if $C=5$, then we have $$5AB=A!+B!+120$$ If $A\le 4$, then the LHS is $\le 100$; and if $A=5$, then the RHS is $>125$. Both cases lead to a contradiction; therefore $C$ can't be $5$.

Hence $C\le 4$. And you seem to have covered all those cases.