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.

Suppose that $p=3k+2$ is prime and 
$$
\left.p\ \middle|\ a^2+ab+b^2\right.\tag1
$$
then, because $a^3-b^2=(a-b)\left(a^2+ab+b^2\right)$, we have
$$
\left.p\ \middle|\ a^3-b^3\right.\tag2
$$
Case $\boldsymbol{p\nmid a}$
Suppose that $p\nmid a$, then $(2)$ says $p\nmid b$. Furthermore,
$$
\begin{align}
a^3&\equiv b^3&\pmod{p}\tag3\\
a^{3k}&\equiv b^{3k}&\pmod{p}\tag4\\
a^{p-2}&\equiv b^{p-2}&\pmod{p}\tag5\\
a^{-1}&\equiv b^{-1}&\pmod{p}\tag6\\
a&\equiv b&\pmod{p}\tag7\\
\end{align}
$$
Explanation
$(3)$: $\left.p\ \middle|\ a^3-b^3\right.$
$(4)$: modular arithmetic
$(5)$: $3k=p-2$
$(6)$: if $p\nmid x$, then $x^{p-2}\equiv x^{-1}\pmod{p}$
$(7)$: modular arithmetic
Then, because of $(1)$ and $(7)$,
$$
\begin{align}
0
&\equiv a^2+ab+b^2&\pmod{p}\\
&\equiv 3a^2&\pmod{p}\tag8
\end{align}
$$
which, because $p\nmid 3$, implies that $p\mid a$, which contradicts $p\nmid a$ and leaves us with
Case $\boldsymbol{p\mid a}$
If $p\mid a$, then $(2)$ says $p\mid b$ and we get
$$
\left.p^2\ \middle|\ a^2+ab+b^2\right.\tag9
$$
