The expression, for `p != 1, (1 + p^(256)) xx (1 + p^(128)) xx (1+ p^(64)) xx (1 + p^(32)) xx (1 + p^(16)) xx (1 + p^8) xx (1 +p^4) (1 + p^2) (1 + p)` is equivalent to :

Options

A. `(1+P^256)/(1+p^128)`
B. `(1-P^512)/(1-p)`
C. `(1-P^256)/(1-p)`
D. `p^255`

Answer

B