import itertools

    def factors(n):
        result = set()
        for i in range(1, int(n**0.5) + 1):
            if n % i == 0:
                result.add(i)
                result.add(n // i)
        return sorted(result)

    def main():
        n = 108
        factors_list = factors(n)
        triplets = []
        for a_idx in range(len(factors_list) - 2):
            a = factors_list[a_idx]
            for b_idx in range(a_idx + 1, len(factors_list) - 1):
                b = factors_list[b_idx]
                for c_idx in range(b_idx + 1, len(factors_list)):
                    c = factors_list[c_idx]
                    if a * b * c == n and a < b < c:
                        triplets.append((a, b, c))
        print("All ways in which three distinct positive integers have a product of 108:")
        for triplet in triplets:
            print(triplet)

    if __name__ == "__main__":
        main()

    -▶ python a.py
    All ways in which three distinct positive integers have a product of 108:
    (1, 2, 54)
    (1, 3, 36)
    (1, 4, 27)
    (1, 6, 18)
    (1, 9, 12)
    (2, 3, 18)
    (2, 6, 9)
    (3, 4, 9)

There's an import for `itertools` which isn't used, just as noted in the article for 4o.

Can someone who knows what it's about say how optimal this version is, compared to other answers?