Show that if $a$ has order $3\bmod p$ then $a+1$ has order $6\bmod p$. Show that if $a$ has order $3\bmod p$ then $a+1$  has order $6\bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $a\bmod p$ is $\phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything.
Thanks!

Note that if $a=1$, $a$ has order $1$. Thus, we can assume $a\ne1$. Furthermore, $p\ne2$ since no element mod $2$ has order $3$. Therefore, $-1\ne1\pmod{p}$.
$$
\begin{align}
(a+1)^3
&=a^3+3a^2+3a+1\\
&=1+3a^2+3a+1\\
&=-1+3(1+a+a^2)\\
&=-1+3\frac{a^3-1}{a-1}\\
&=-1
\end{align}
$$
Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn). 
$$
\begin{align}
(a+1)^2
&=a^2+2a+1\\
&=a+(a^2+a+1)\\
&=a+\frac{a^3-1}{a-1}\\
&=a\\
&\ne1
\end{align}
$$
Therefore, $(a+1)^2\ne1$.
Thus, $(a+1)$ has order $6$.
