Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will have a factor of 4
One of them will have a factor of 3
One of them will have a factor of 2
=> 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.
Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated
Thank you

We know from here or here, the product $n$  consecutive integers is divisible by $n!$ for integer  $n>0$
$$(a-2)(a-1)a\cdot a(a+1)(a+2)$$
$$=(a-2)(a-1)a(a+3-3)(a+1)(a+2)$$
$$=\underbrace{(a-2)(a-1)a(a+1)(a+2)(a+3)}_{\text{ the product of } 6\text{ consecutive integers }}-3\underbrace{(a-2)(a-1)a(a+1)(a+2)}_{\text{ the product of } 5\text{ consecutive integers }}$$
We could divide $a$ as $a-3+3$ as well.
