Tricky one. You can solve it unfolding the expression using the (a-b)(a+b property or recognize a pattern

of 2^{16}-1 notice that the smallest root of two is 2^2 (essentially we do have multiple twos)

So at the core we do have $(2^2-1) = (2-1)(2+1)$ that is $1*(2+1)$

In the end we do have for QA $(2+1)(2^2+1)(2^4+1)(2^8+1)$

Which is = to QB

C is the answer

Very tricky and good question Thanks!!!Carcass

Is there anyone who can break this question down? I really can't wrap my head around it.

If you use

$a^2-b^2=(a-b)(a+b)$

Therefore

$(2^8-1^8)(2^8+1^8)$

$(2^4-1^4)(2^4+1^4)(2^8+1^8)$

$(2^2-1^2)(2^2+1^2)(2^4+1^4)(2^8+1^8)$

$(2-1)(2+1)(2^2+1^2)(2^4+1^4)(2^8+1^8)$

$(2+1)(2^2+1^2)(2^4+1^4)(2^8+1^8)$

C is the answer

Thank you. This was very helpful