First, when $p=2$, $p^2+8=12$ is not a prime. Next, when $p=3$, both $p^2 + 8=17$ and $p^3 + 8p+2=53$ are prime.

Therefore, the conclusion holds when $p\le 3$. Now, let's show that when $p > 3$, $(p^2+8)$ cannot be a prime.

In this case, given $p$ is prime, we have have $p\not\equiv 0\pmod{3}$. However, $$p^2 + 8\not\equiv 0\pmod{3}\implies p^2\not\equiv 1\pmod{3}\implies p\not\equiv\pm 1\pmod{3}$$

This is impossible because one of $p\equiv 0\pmod{3}$ and $p\equiv\pm 1\pmod{3}$ must hold.