<s>[INST] 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
 [/INST] If one of $a,b,c$ is divisible by $7,abc\equiv0\pmod 7$
Else
$n^3\equiv \begin{cases} 1 &\mbox{if } n \equiv 1,2,4\pmod 7 \\
-1 & \mbox{if } n \equiv 3,5,6\pmod 7 \end{cases} \pmod 7$
Observe that for no combination of $a,b,c$ $$a^3+b^3+c^3\equiv0\pmod 7$$
$$\implies a^3+b^3+c^3\equiv0\pmod 7\implies 7\text{ must divide at least one(or all three) of } a,b,c $$
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