<s>[INST] Prove that : $a,b,c\in\mathbb{Z}$ then : $abc\equiv 0\pmod{3} \implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$ Prove that : 
$a,b,c\in\mathbb{Z}$ then : 
$$abc\equiv 0\pmod{3} \implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
My try : 
Let $a\not\equiv 0\pmod{3}$
Then : 
$$a\equiv 1,2\pmod{3}$$
So : 
$$a^{3}\equiv ±1\pmod{9}$$ 
Similarly :
$$b^{3}\equiv ±1\pmod{9}$$
$$c^{3}\equiv ±1\pmod{9}$$
$$\implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
Is my solution correct ?? 
And we can generalized or no ?
$$abc\equiv 0\pmod{3} \iff a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
 [/INST] As pointed in the comments , the equivalence is false but it's converse is always true . $$a^3+b^3+c^3 \equiv 0 \mod9\implies abc\equiv 0\mod3$$
Since a cube can be written as $x^3\equiv 0,1,8 \mod 9$  , the only way to add these residues to be divisible by $9$ is $(0,0,0)$ and $(0,1,8)$ and their permutations. 
In either the case we get at least one cube divisible by $9$ implying that the number is divisible by $3$ and the conclusion follows.

The problem with your attempt is that you assumed $$a^3 \equiv b^3\equiv c^3 \equiv \pm 1\mod 9$$  which implies that $3\not\mid a,b,c$ , which means that $abc\equiv 0\mod 3$ is false.
</s>