The question says:

Prove that if $a$ is of order $3$ modulo a prime $p$, then $1+a+a^2equiv 0 pmod p$. Moreover, $a+1$ is of order $6$.

For the First Part:

The typical idea is to start with $a^3 equiv 1 pmod p to a^3 -1 equiv 0 pmod p$. Factoring the term on the left hand side, the rest is straightforward.

However, I need to check the following idea:

$$1+a+a^2 equiv a^3+a^2+a equiv a(1+a+a^2)equiv a^2(1+a+a^2)$$
$$equiv a^3(1+a+a^2) equiv 0 pmod p$$

Since, by the hypothesis, $a^3$ cannot be zero modulo the prime $p$, the desired result holds.

If this idea holds true, it can be generalized to the following result:

if $a$ is of order $k$, then, modulo prime $p$, then $1+a+a^2+cdots + a^{k-1}$ is divisible by $p$.

Is it??

For the Second Part:

I can see that:

$$1+a+a^2 equiv 0 to 1+a equiv -a^2 pmod p$$

However, does this implies anything? Given that $a$ is of order $3$, but what about $-a$??

Please Help, and Thanks in advance,,

For the second part, taking from the first part that $1+a = -a^2pmod pimplies (1+a)^6= (-a^2)^6 = a^{12}pmod p= (a^3)^4 =1^4pmod p=1pmod p$ . Thus $1+a$ is of order $6$ as claimed. And if $a$ is of order $3$ then $(-a)^3 = -a^3 = -1 = p-1pmod p$. And from this we have: $(-a)^6 = ((-a)^3)^2 = (-1)^2 = 1pmod p$. So $-a$ is of order $6$ as well.