To solve this problem, we need to find the number of three-digit numbers with distinct digits such that the product of the digits is the cube of a positive integer.

- We know that a cube number is a number that is obtained by multiplying an integer by itself twice. So, let's consider the cube numbers less than 10:
- The cube of 1 is 1.
- The cube of 2 is 8.
- The cube of 3 is 27.
- The cube of 4 is 64.
- The cube of 5 is 125.
- The cube of 6 is 216.
- The cube of 7 is 343.
- The cube of 8 is 512.
- The cube of 9 is 729.

- We can see that there are 5 cube numbers less than 10: 1, 8, 27, 64, and 125.

- The product of the digits in a three-digit number can range from 1 (if all the digits are 1) to 9 x 8 x 7 = 504 (if the digits are 9, 8, and 7).

- We need to find three-digit numbers where the product of the digits is one of the cube numbers less than 10.

- Let's consider each cube number and find the possible combinations of three-digit numbers with distinct digits that have that cube number as the product of the digits:

- For the cube number 1, the only possible combination is 111. So, we have 1 three-digit number for this cube number.

- For the cube number 8, the possible combinations are 248 and 824. So, we have 2 three-digit numbers for this cube number.

- For the cube number 27, the possible combinations are 369, 693, and 936. So, we have 3 three-digit numbers for this cube number.

- For the cube number 64, there are no possible combinations since the maximum product of three distinct digits is 504.

- For the cube number 125, the possible combinations are 125 and 512. So, we have 2 three-digit numbers for this cube number.

- Therefore, the total number of three-digit numbers with distinct digits such that the product of the digits is the cube of a positive integer is 1 + 2 + 3 + 0 + 2 = 8.

- The correct answer is option D: 8.