Solution:Statement 1
n³ - n = n(n-1)
= n (n² - 1²)
= n (n - 1) (n + 1)
= (n - 1) n (n + 1)Here, (n - 1) * n(n + 1) are the product of three consecutive integer which is always divisible by 2 and 3 i.e., 6
Statement 2
n⁵ - n
n(n⁴ - 1)
= n(n² - 1) (n² + 1)
= n(n - 1) (n + 1) (n² + 1)
= (n - 1) n (n + 1) (n² - 4 + 5)
= (n - 2) (n - 1) (n) (n + 1) (n + 2)
+ 5(n - 1) n (n - 1)
= Five consecutive integer multiple
5(n - 1) n (n - 1)
So, it is divisible by 5! i.e.,5
Statement 3
n⁵ - 5n³ + 4n
= n(n² - 4) (n² - 1)
= (n - 2) (n - 1) n (n + 1) (n + 2)
= Multiple of 5 consecutive integer
So, it is divisible by 5! = 120