To show that $a^3, b^3, c^3$ give different residues modulo $p$, where $p = ab + bc + ac$ is a prime number and $a, b, c$ are distinct integers, we can proceed as follows:

1. Given Conditions and Polynomial Formation:
Since $a, b, c$ are distinct and $p = ab + bc + ac$ is a prime number, we know that $a, b, c$ are roots of the polynomial:
$$P(x) = (x - a)(x - b)(x - c) = x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc$$
In the field $\mathbb{F}_p$, this polynomial can be written as:
$$P(x) = x^3 - (a+b+c)x^2 + px - abc$$
Since $p = ab + bc + ac$, the polynomial simplifies to:
$$P(x) = x^3 - (a+b+c)x^2 + 0 \cdot x - abc = x^3 - (a+b+c)x^2 - abc$$

2. **Non-zero Sum Modulo $p$:**
Since $a, b, c$ are distinct, $a + b + c$ is non-zero modulo $p$. This is because if $a + b + c \equiv 0 \pmod{p}$, then $a, b, c$ would not be distinct roots of the polynomial $x^3 - abc$ modulo $p$.

3. **Distinct Cubes Modulo $p$:**
Consider the function $f(x) = x^3$ in $\mathbb{F}_p$. We need to show that $a^3, b^3, c^3$ are distinct modulo $p$. Suppose for contradiction that $a^3 \equiv b^3 \pmod{p}$. Then:
$$a^3 - b^3 \equiv 0 \pmod{p}$$
This implies:
$$(a - b)(a^2 + ab + b^2) \equiv 0 \pmod{p}$$
Since $p$ is a prime and $a \neq b$, it must be that:
$$a^2 + ab + b^2 \equiv 0 \pmod{p}$$
Similarly, if $b^3 \equiv c^3 \pmod{p}$ or $a^3 \equiv c^3 \pmod{p}$, we would get:
$$b^2 + bc + c^2 \equiv 0 \pmod{p} \quad \text{or} \quad a^2 + ac + c^2 \equiv 0 \pmod{p}$$
However, since $a, b, c$ are distinct and $a + b + c \not\equiv 0 \pmod{p}$, these equations cannot hold simultaneously. Therefore, $a^3, b^3, c^3$ must be distinct modulo $p$.

$\blacksquare$