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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 $3 \mid (a+b+c+d)$?

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$x^3\equiv x\bmod3$, 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 $3|(a+b+c+d)$.

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    $\begingroup$ Or you can see that $x^3 \equiv x \pmod{3}$ by Fermat's Little Theorem. $\endgroup$ – user19405892 Jan 7 '17 at 3:17
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    $\begingroup$ @user19405892 Normally, I wouldn't advice simple brute force methods, but $$0^3\equiv0\bmod3$$ $$1^3\equiv1\bmod3$$ $$2^3\equiv2\bmod3$$ $\endgroup$ – AlgorithmsX Jan 7 '17 at 3:26

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