Solution:

Let $r$ and $s$ be two different integral roots of $x^{3} + a x^{2} + b x + c = 0$; that is, $r^{3} + a r^{2} + b r + c = 0$ and $s^{3} + a s^{2} + b s + c = 0$. Since $p$ divides $a$, $b$, and $c$, it follows that $p$ divides both $r^{3}$ and $s^{3}$. Being prime, $p$ divides $r$ and $s$.

Subtracting the above equations involving $r$ and $s$, we get
$$r^{3} - s^{3} + a(r^{2} - s^{2}) + b(r - s) = 0, \text{ or } (r - s)\left(r^{2} + r s + s^{2} + a(r + s) + b\right) = 0$$
Since $r \neq s$, the last equation becomes
$$r^{2} + r s + s^{2} + a(r + s) + b = 0$$
Because the terms (other than $b$) are divisible by $p^{2}$, the last equation forces $p^{2}$ to divide $b$.

Finally, the terms (other than $c$) of $r^{3} + a r^{2} + b r + c = 0$ are divisible by $p^{3}$, it follows that $p^{3}$ divides $c$.