Final answer:
The product (a-b)(a-c)(b-c) of three distinct prime numbers greater than 3 is divisible by 48, due to each factor being even (thus divisible by 2) and one being divisible by 4 and another being a multiple of 3.
Step-by-step explanation:
We are given three prime numbers greater than 3, and they are represented as (a-b)(a-c)(b-c). Since all three primes are greater than 3, they must be odd. When we subtract one odd number from another, the result is always even. This means that each of the factors, (a-b), (a-c), and (b-c), is an even number. Moreover, because the difference between any two distinct prime numbers greater than 3 is at least 2, each of these even numbers is divisible by at least 2.
Next, we observe that one of these even numbers must be divisible by 4. If we consider the differences between our prime numbers a, b, and c, at least one of these must be even and not just twice an odd number (which would yield a result divisible by 2, but not necessarily by 4). Since the primes are distinct and greater than 3, one of the differences will be divisible by 4. This addresses divisibility by 2 and 4.
Additionally, since we have three distinct, odd primes, at least one of the differences (a-b), (a-c), or (b-c) must be a multiple of 3. As prime numbers are more than 3, they can't be multiples of 3, so the only way for their difference to be a multiple of 3 is if it spans across a multiple of 3, which necessitates that it is indeed divisible by 3.
Therefore, the product (a-b)(a-c)(b-c) is divisible by 2 from each factor, 4 from the larger even factor, and 3 from the factor that spans the multiple of 3. The product of these divisors is 2*2*2*2*3, which equals 48. Thus, (a-b)(a-c)(b-c) is always divisible by 48.