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 $n$

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 ("$c>6$").
