Here is one scenario that seems to work:
Divisors of 24 =(1, 2, 3, 4, 6, 8, 12, 24) =8 divisors.
proper product =(1, 2, 3, 4, 6, 8, 12)=13,824. Notice that I did not include 24 itself.
24^3 =13,824=24^(8*1 + 10) / 6, where:
x=8, a =1, b=10 and c=6.
GCD{1, 6, 10} = 1
a + b + c =1 + 10 + 6 = 17
Note: Also it seems to work for 60:
Divisors of 60 =(1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) = 12
Proper product( (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30) =777,600,000. Notice again I did include 60 itself.
60^5 =777,600,000 =60^(12*1 + 13) / 5, where:
x =12, a = 1, b = 13, and c =5
GCD, or GCF[1, 5, 13] = 1
a + b + c =1 + 13 + 5 = 19.
P.S. I believe there are infinite solutions to your question.