Proving rigorously that if $3\mid(a+b)$ then $3\mid(a^3+b^3)$ using divisibility definition Let $a,b\in \mathbb Z$. Prove rigorously using divisibility definition that if $3\mid(a+b)$ then $3\mid(a^3+b^3)$
After a bit of algebra I get that
$$3\overset{?}{\mid}(a+b)^3-3ab(a+b)$$
So now how do I justify that it's divisible by $3$? Can I show that both expressions are divisible by $3$ separately? Then what do I mention to justify the sum of divisors is divisible?

If $3|(a+b)$ then $3|((a+b)^3-3a^2b-3ab^2)$ so $3|a^3+b^3$.
