Question 251228
using this {{{(x^2-y^2)=(x+y)(x-y)}}}

we get {{{(16n^4-1)=(4n^2-1)(4n^2+1)}}}

4n^2-1 can be further divided into (2n+1)(2n-1)

so {{{(16n^4-1)=(2n-1)(2n+1)(4n^2+1)}}} Ans.