Bunuel wrote:

\((3^2+1)(81^2+1)(9^2+1)(3^2−1)=\)

A. \(81^4+1\)

B. \(9^9−1\)

C. \(3^{16}−1\)

D. \(3^{32}−1\)

E. \(3^{96}+1\)

Property : a^2-b^2= (a+b)(a-b)

So, (3^2+1)(3^2−1) = 3^4-1

The expression becomes :

(3^4-1)(3^8+1)

(3^4+1)Expression= (3^8+1)(3^8-1)

or, (3^16-1)

Answer C Simple as that!

Or through another route-

The expression becomes : (3^4-1)(81^2+1)(9^2+1)

Since (3^4-1) = (9^2-1)

Expression=

(9^2-1)(81^2+1)

(9^2+1)or, (81^2+1)(81^2-1)

or, (81^4-1)

or, (3^16-1)

C again

_________________

It is not who I am underneath but what I do that defines me.