Show that $a^3 - b^3 = c! - 18$ does not have a solution 
Let $a, b,$ and $c$ be positive integers and $c \gt 6$.
Show  that the equation $$a^3 - b^3 = c! - 18$$ does not have a solution for all positive integers $a, b,$ and $c$.

What I have realized so far is that if $a^3 - b^3 = c! - 18$, then it must also be true that $a^3 -b^3 \equiv c! - 18 \pmod n$ for all $n \geq 2$.
That would mean if I can show that there exists an $n$ so that $$a^3 - b^3 \not\equiv c! - 18 \pmod n$$ for all relevant $a, b,$ and $c$ then I have also shown that $$a^3 - b^3 \neq c! - 18$$ for all relevant $a,b,$ and $c$.
The only problem that I have now is that I do not know how to find a suitable $n$

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.
