Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.  
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$
So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.  
I do not know what to do next. Please help. Thank you.

You have shown that $a^3 \equiv b^3 \pmod p$. Remark that $(3, p-1)=1$ because $p=3k+2$. Thus we can write $3m + (p-1)n = 1$ for some integers $m, n$. Use this to show that $a\equiv b \pmod p$, so that $a^2+ab+b^2 \equiv 3a^2 \equiv 3b^2 \equiv 0 \pmod p$, and conclude from there. 
(P.S. I have to commend you for posting your work!)
