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!

ord_$pa=3\iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$
But $p\not\mid (a-1)$ as ord_$pa=3 $
$\displaystyle\implies a^2+a+1\equiv0\pmod p$
$\displaystyle\iff a(a+1)\equiv-1\pmod p$
Method $1:$
$\implies a^3(a+1)^3\equiv-1\pmod p\implies (a+1)^3\equiv-1$ and $(a+1)^6\equiv1$
$\implies $ord_$p(a+1)\mid 6$ but does not divide $3$
If ord_$p(a+1)\mid 2, (a+1)^2\equiv1\pmod p$ and $a^2(a+1)^2\equiv1\pmod p\implies a^2\equiv1$ which is impossible as ord$_pa=3$
$\implies $ord$_p(a+1)=6$
Method $2:$
$\displaystyle a(a+1)\equiv-1\pmod p\iff a+1\equiv (-a)^{-1}\pmod p$
Again as $a^3\equiv1\pmod p, (-a)^3=-a^3\equiv-1\pmod p\implies (-a)^6\equiv1\implies $ord$_p(-a)=6$
Using this,  ord$_p\{(-a)^{-1}\})=$ord$_p(-a)=6$
