If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$. 
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is  divisible by $7$.

Here it seems Fermat's theorem  has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's tedious.
Any ideas for a much shorter solution?

By Fermat's theorem $x^6\equiv1\pmod7$ holds for all $x$ coprime to $7$, so $(x^3)^2\equiv1\pmod7$ hence
$$(x^3)^2-1=(x^3-1)(x^3+1)\equiv0\pmod7$$
and hence $x^3\equiv\pm1\pmod{7}$. 
