Show that $a^3 - b^3 = c! - 18$ does not have a solution

You may want to try your luck with $\bmod 7$. You should know, or easily show, that all cubes are $\in\{0,1,6\}\bmod 7$ forcing $a^3-b^3\in\{0,1,2,5,6\}$. But for $c\ge 7$ you find $c!-18$ fails to meet this qualification (I will let you figure that out). This does not generally work for $c<7$, but that was excluded from the problem statement ("$c>6$").

Consider modulo $3$. When $c$ is sufficiently large, arbitrarily large (in particular, $>2$) powers of $3$ divide $c$. Suppose $c$ is sufficiently large in this sense, and suppose $a,b$ existed so that the equation is satisfied. The RHS is $0$ modulo $3$, so we have $a^3=b^3$ mod $3$, which you can easily check implies $a=b$ mod $3$. If $a=b=0$ mod $3$, then the LHS is divisible by $27$, but the RHS is not (since $3\mid c!/9$ but $3\nmid 2$). Suppose $a=b=1$ mod $3$, then set $a=3n+1$, $b=3m+1$. So $a^3-b^3=27(n^3+n^2-m^3-m^2)+9(n-m)=0$ mod $9$. Again, we have a contradiction with the fact that the RHS is not divisible by $9$. Simlarly, then $a=3n-1$ and $b=3m-1$, we have $a^3-b^3=(3n-1)^3-(3m-1)^3=27(n^3-n^2-m^3+m^2)+9(n-m)=0$ mod $9$, and the conclusion follows.