If $a$ has order 3 $\pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$ There is a little proposition left  to the readers in my number theory book, and I'm not sure how to answer this in a proper way:

If $a$ has order $3 \pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$.

Some hints?
 A: Work in the residue class ring $\bf Z/p\bf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1\ne 0$ if $a$ has order $3$.
For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?
A: $\!\bmod p\!:\,\ a^3\equiv 1, a\not\equiv 1\,$ so $\,0\equiv a^3-1\equiv (a-1)(a^2+a+1)\,\Rightarrow\,a^2+a+1\equiv 0$ 
Thus $\ \ 1+a\equiv -a^2.\ $ To compute $\,(1+a)^n\,$ we power this congruence:
Thus $\,(1+a)^6\equiv\, a^{12}\equiv 1,\ $ by $\,a^3\equiv 1$ 
But $\ \ \color{#0a0}{(1+a)^2}\equiv\ a^4\, \equiv\, a\, \color{#0a0}{\not\equiv 1}\,$ by hypothesis
and $\ \,  \color{#0a0}{(1+a)^3}\!\equiv\! -a^6\!\equiv\! -1 \color{#0a0}{\not\equiv 1}\,$ by $\,p\neq 2\, $ (where only order $1$ is possible)
Thus $\,1+a\,$ has order $\,6\,$ by the Order Test below, with $\,\color{#0a0}{p = 2,3}$
Order Test $\ \ \ \,a\,$ has order $\,n\ \iff\  a^{\large n} = 1\ $ but $\ \color{#0a0}{a^{\large n/p} \neq 1}\,$ for every prime $\,p\mid n$ 
Proof $\,\ (\Leftarrow)\,\ $ Let $\,a\,$ have $\,\color{#c00}{{\rm order}\ k}.\,$ Then $\,k\mid n\,$ (proof). $ $  If $\:k < n\,$ then $\,k\,$ is proper divisor of $\,n\,$ so by unique factorization $\,k\,$ arises by deleting at least one prime $\,p\,$ from the prime factorization of $\,n,\,$ so $\,k\mid n/p,\,$ say $\, kj = n/p,\ $ so $\ \color{#0a0}{a^{\large n/p}} = (\color{#c00}{a^{\large k}})^{\large j} = \color{#c00}1^{\large j} = \color{#0a0}1\,$ contra $\rm\color{#0a0}{hypothesis}$. So $\,k=n.$ $\, (\Rightarrow)\ $ Clear
A: As $a$ has order $3$, $a^3\equiv 1 \pmod p$. Hence $a^3-1 \equiv 0 \pmod p$.
So $a^3-1=(a-1)(1+a+a^2)\equiv 0 \pmod p$.
Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.
A: How $a$ has order $3$, then $a\not\equiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 \equiv 0$ (mod $p$). As $\mathbb{Z}_p$ is a field and $a-1\neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that 
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1\Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1\Rightarrow 1+2a+a^2=1\Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1\Rightarrow (1+a)(1+2a+a^2)=1\Rightarrow (1+a)a=1\Rightarrow a+a^2=1\Rightarrow 1+a+a^2=2\Rightarrow 0=2$, so $p=2$. But $\mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
