If none of $a,b,c$ are divisible by 3 then we get: $$a^3 \equiv \pm 1 \pmod{9} \Rightarrow \sum a^3 \equiv -3,-1,1,3 \pmod{9}$$But this contradicts $9 \vert \sum a^3$ so we must have $3 \vert abc$. Similarly if all of $a,b,c$ are odd then we would have $\sum a^3 \equiv 1 \pmod{2}$ which is a contradiction so $2 \vert abc$. Combining these two results gives: $$6 \vert abc$$

We just have to show that at least one of the numbers $a$, $b$ and $c$ is a multiple of $2$ and at least one the numbers $a$, $b$ and $c$ is a multiple of $3$. Indeed, all three number cannot be odd, otherwise, $a^3+b^3+c^3$ is odd and would not be a multiple of $18$. Therefore, at least one of the numbers $a$, $b$ and $c$ is a multiple of $2$. On the other hand, for Fermat's lettle theorem, we know that $$0\equiv a^3+b^3+c^3\equiv a+b+c\pmod 3.$$So, we have that $3\mid a+b+c$, therefore, $a\equiv b\equiv c\pmod 3$ or $a$, $b$ and $c$ are $1$, $2$ and $3$ in mod $3$ (in some order). If the second is fulfilled, the problem is solved; but if the first is fulfilled, then there is numbers $x$, $y$ and $z$ such that $a=3x+r$, $b=3y+r$ and $z=3z+r$, where $r$ is $1$ or $2$. Then, $$a^3+b^3+c^3\equiv 3r^3\pmod 9,$$but $r$ is $1$ or $2$, then $3r^3\equiv 3\pmod 9$ o $3r^3\equiv 6\pmod 9$, which implies that $a^3+b^3+c^3$ is not a multiple of $18$, and this is absurd. Therefore, at least one of the numbers $a$, $b$ and $c$ is a multiple of $3$. Finally, the product $abc$ is a multiple of $2$ and $3$, then $abc$ is a multiple of $2\times 3=6$.

If $a,b,c$ are all divisible by $3,$ then obviously, this is true. Now, if none of $a,b,c$ are divisible by $3,$ we have that $a^3, b^3, c^3 \equiv \pm 1\pmod{9}$, but this yields, a contradiction. Hence, if $2,$ of $a,b,c$ are divisible by $3,$ we also have a contradiction. So, we have that exactly one of $a,b,c$ are divisible by $3.$ Note, that these cubes $-1, 0, 1\pmod{9}$, we have that $3,$ is a divisor of the product. Note, that $a^3+b^3+c^3$, is even, and if $a,b,c$ are all odd, then we have a contradiction. Hence, one of $a,b,c$ must be even, hence, $6|abc.$