Correct Answer: $\frac{x^{256}-1}{x-1}$

Solution : Given: $(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)$
Multiplying $(x-1)$ in both numerator and denominator,
$\frac{(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)(x-1)}{(x-1)}$
Using identity: $(a^2-b^2)=(a-b)(a+b)$,
= $\frac{(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x^2-1)}{(x-1)}$
= $\frac{(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^4-1)}{(x-1)}$
= $\frac{(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^8-1)}{(x-1)}$
= $\frac{(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{16}-1)}{(x-1)}$
= $\frac{(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{32}-1)}{(x-1)}$
= $\frac{(x^{128}+1)(x^{64}+1)(x^{64}-1)}{(x-1)}$
= $\frac{(x^{128}+1)(x^{128}-1)}{(x-1)}$
= $\frac{(x^{256}-1)}{(x-1)}$
Hence, the correct answer is $\frac{(x^{256}-1)}{(x-1)}$.