Prove that $3 \mid (a+b+c+d)$ 
Given $a,b,c,d \in \mathbb{Z}$ satisfying $a^3+b^3 = 2(c^3-8d^3)$, prove that $3 \mid (a+b+c+d)$.

I first factorized $a^3+b^3$ to get $a^3+b^3 = (a+b)(a^2-ab+b^2)$. I wasn't sure how to use the right-hand side to get $a+b+c+d$. How can we prove that $3 \mid (a+b+c+d)$?

$x^3\equiv x\bmod3$, which you can prove by trial and error in this case. This should reduce it down to $a+b\equiv2c-d\bmod3$. Adding $c+d$ to both sides yields $a+b+c+d\equiv3c\equiv0\bmod3$, which means $3|(a+b+c+d)$.
