Answer: To find the remainder when dividing the sum of cubes of factorials by 144, we need to simplify the expression and then find the remainder.
First, let's calculate the cubes of the factorials:
(1!)^3 = 1^3 = 1
(2!)^3 = 2^3 = 8
(3!)^3 = 6^3 = 216
...
We can notice a pattern that for each factorial (n!), its cube will be (n!)^3 = n^3 * (n!)^2.
Now, let's calculate the squares of the factorials:
(1!)^2 = 1^2 = 1
(2!)^2 = 2^2 = 4
(3!)^2 = 6^2 = 36
...
Again, we can observe a pattern that for each factorial (n!), its square will be (n!)^2 = n^2 * (n-1)^2 * (n-2)^2 * ... * 2^2 * 1^2.
Now, let's calculate the remainder when dividing each term by 144:
(1!)^3 % 144 = 1 % 144 = 1
(2!)^3 % 144 = (2^3 * (2!)^2) % 144 = (8 * (2^2 * (2-1)^2)) % 144 = (8 * (4 * 1)) % 144 = 32 % 144 = 32
(3!)^3 % 144 = (216 * (3!)^2) % 144 = (216 * (3^2 * (3-1)^2 * (3-2)^2)) % 144 = (216 * (9 * 4 * 1)) % 144 = 7776 % 144 = 72
...
Now, let's continue this process until (432!)^3:
(432!)^3 % 144 = (432^3 * (432!)^2) % 144 = (432^3 * (432^2 * (432-1)^2 * ... * 2^2 * 1^2)) % 144
Since all the terms (432^2, (432-1)^2, ..., 2^2, 1^2) are multiples of 144, they won't affect the remainder. So, we only need to find the remainder of 432^3 divided by 144.
Now, calculate the remainder:
432^3 % 144 = (432^2 * 432) % 144 = (186624 * 432) % 144 = 802816 % 144 = 0
Therefore, the remainder when dividing the given expression by 144 is 0 (option A).