Question about modular arithmetic and divisibility If $$a^3+b^3+c^3=0\pmod 7$$
Calculate the residue after dividing $abc$ with $7$
My first ideas here was trying the 7 possible remainders and then elevate to the third power
$$a+b+c=x \pmod 7$$
$$a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc=x^3\pmod 7$$
$$3(a+b+c)(ab+bc+ac)-3abc=x^3 \pmod 7$$
If I replace $x=0$ the result is immediate, $abc=0 \pmod7$. But with $x=1$ 
$$3(7n+1)(ab+bc+ac)-3abc=x^3 \pmod 7$$
$$3(ab+bc+ac)-3abc=x^3 \pmod 7$$
And there is nothing more to simplify. I know the LHS is a multiple of $3$, but what can i do with that? Is it necessary that $x^3$ or $7-x^3$ is a multiple of $3$? Any help is greatly appreciated

HINT: What are the possible cubes in modulo 7? Hence what combinations of these cubes allow for $a^3+b^3+c^3=0$ (mod 7). From this $abc$ (mod 7) should be clear.
