If $p$ and $p^2+8$ are both prime numbers, then $p^3+4$ is prime. In the book, "Elementary Number Theory - 6th Edition" written by David M. Burton, I have a question. The problem is
If $p$ and $p^2+8$ are both prime numbers, prove that $p^3+4$ is also prime.
(p. 58 problem 3.3 #20)
I don't know how to prove this.
Thank you very much if you solve this question.

Hint If $ p \neq 3$ prove that $p^2+8 \equiv 0 \pmod {3}$. 
Thus $3 |p^2+8$. Can you finish it from here?
P.S. You can prove the first claim without modular arithmetic by observing that 
$$p^2+8=p^2-1+9=(p-1)(p+1)+9$$
Both terms in the sum are multiple of $3$.
