996. Find all cases when a natural number has 8 different divisors.
$\triangle$ Based on the statement of problem 995 (with the same notation), we obtain:

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{n}+1\right)=8
$$

On the other hand, let's find all factorizations of the number 8 into a product of natural numbers:

$$
8=1 \cdot 8=2 \cdot 4=2 \cdot 2 \cdot 2
$$

Accordingly, we will consider three cases.

1) In the first case

$$
\alpha_{1}+1=1, \quad \alpha_{2}+1=8
$$

from which $\alpha_{1}=0, \alpha_{2}=7$. The desired number $a$ takes the form $a=p^{7}$, where $p$ is a prime.

2) In the second case

$$
\alpha_{1}+1=2, \quad \alpha_{2}+1=4
$$

Then

$$
\alpha_{1}=1, \quad \alpha_{2}=3, \quad a=p_{1} p_{2}^{3}
$$

where $p_{1}$ and $p_{2}$ are different prime numbers.

3) In the third case

$$
\alpha_{1}+1=2, \quad \alpha_{2}+1=2, \quad \alpha_{3}+1=2
$$

Therefore,

$$
\alpha_{1}=\alpha_{2}=\alpha_{3}=1, \quad a=p_{1} p_{2} p_{3},
$$

where $p_{1}, p_{2}$ and $p_{3}$ are different prime numbers.

Answer: $a=p^{7}$, where $p$ is a prime; $a=p_{1} p_{2}^{3}$, where $p_{1}$ and $p_{2}$ are different prime numbers; $a=p_{1} p_{2} p_{3}$, where $p_{1}, p_{2}$ and $p_{3}$ are different prime numbers.