Prove that (4n+1)^2(4n-1)^2 is a multiple of 8 for all integers of n.

thanks.

i think  (4n+1)^2(4n-1)^2 isnt a multiple of 8 for all integers of n because:

(4n + 1)²(4n - 1)²

= [(4n + 1)(4n - 1)]²

= (16n² - 1)²

= 16².n².n² - 2.16.n² + 1

= 8n²(32n² - 4) + 1

can see 8n²(32n² - 4) is a multiple of 8 but 1 isnt a multiple of 8

=> (4n + 1)²(4n - 1)² isnt a multiple of 8 for all integers of n.

Step-by-step explanation:

incomplete question. how do you expect anyone to you when you do not post the question in its entirety?

i think da answer is

d.square root of 3