If integer C is randomly selected from 20 to 99, inclusive.

To be divisible by 12, c³-c must be a multiple of 3 and 4.

c³-c = c(c²-1) = c(c-1)(c+1).
c-1, c, and c+1 are three consecutive integers.

Of every three consecutive integers, exactly one is a multiple of 3.
Thus, (c-1)(c)(c+1) is a multiple of 3.

The question is whether (c-1)(c)(c+1) will be a multiple of 4.
If c is odd, then (c-1)(c)(c+1) = (even)(odd)(even).
In this case, the product will be a multiple of 4, since it includes two even factors.

If c is a multiple of 4, then (c-1)(c)(c+1) = (odd)(multiple of 4)(odd).
In this case, the product will be a multiple of 4, since one of its factors is a multiple of 4.

If c is an even integer that is NOT a multiple of 4, then (c-1)(c)(c+1) = (odd)(even non-multiple of 4)(odd).
In this case, (c-1)(c)(c+1) will NOT be a multiple of 4.

Between 20 and 99:
The total number of integers = 80.
Since half are even, the number of even integers = 40.
Since every other even integer is a multiple of 4, the number of even integers that are NOT multiples of 4 = 20.
Thus, P(even non-multiple of 4) = 20/80 = 1/4.

Since any other value for c will yield a multiple of 4 for (c-1)(c)(c+1), we get:
P(c³-c will be a multiple of 12) = 1 - 1/4 = 3/4.