Take a number N. Find N such that P(n)=n^3 where P(n) is a function that multiplies ALL the factors of that number.
All of the listed numbers are answers that I found
What to look for is when all the factors of the number without the number itself equal the number squared. The first thing to note is that the solutions are not prime, because then P(n)=n. Also, that first example (12) has 6 factors (1,2,3,4,6,12) Other numbers with 6 factors are multiplication of a prime number and a square of a prime, because 101*9 is 909, and factors of 909 are 1, 3, 9, 101, 303, 909, and since 3x303 and 9x101 both equal 909, 909 is a valid solution. So now there are an infinite amount of numbers that have 6 factors. Although this examples listed are arbitrary, the math holds.
If you think about it, and seeing that I don't have as great of math skill as the people before me, the factors are the key. If N has 6 factors, then P(n) is equal to n^3. NOT SAYING THERE ARE NO OTHER SOLUTIONS. So if there are other solutions, I do not know of them. All of the solutions that you provided have 6 factors, so therefore, yes, all those are solutions.
And all others variations of this puzzle, for instance instead of cubing N, we could square N, and then all numbers with 4 factors (cubes of primes, 2 primes multiplied by each other, et cetera) would be a solution, or leave it alone (raising it to a power of 1), then we get all the primes.
I should mention a growing pattern that for any integer K and any integer N, if P(N)=N^K, then N has 2K factors, guaranteed.