# 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

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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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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.

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