Question

$(2r-1)(2r+1)(4r^2+1)$

Answer 1

(2r−1)(2r+1)(4r^2+1)

=(4r^2-1)(4r^2+1)

=16r^4-1

Answer 2

Answer:16r^4-1

Given that : (2r-1)(2r+1)(4r^2+1)

Now it can be written as [(2r-1)(2r+1)](4r^2+1)

first two terms are in the form of (a-b)(a+b)

we know that (a-b)(a+b)=a^2-b^2

now we have ,

[(2r)^2-(1)^2](4r^+1)

(4r^2-1)(4r^2+1)

again this is in the form of (a+b)(a-b)

so , we get (4r^2)^2-1^2= 16r^4-1