Question : What is the simplified value of $(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})?$

Option 1: $(x^{64}+\frac{1}{x^{64}})$

Option 2: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x^2+\frac{1}{x^2})}$

Option 3: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x+\frac{1}{x})}$

Option 4: $\frac{(x^{32}-\frac{1}{x^{32}})}{(x+\frac{1}{x})}$

Correct Answer: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x^2+\frac{1}{x^2})}$

Solution : Given:
$(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})$
$=(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^2-\frac{1}{x^2})(x^{16}+\frac{1}{x^{16}})(x^{4}+\frac{1}{x^{4}})$
Multiplying it by $\frac{(x^2+\frac{1}{x^2})}{(x^2+\frac{1}{x^2})}$, we get,
$= \frac{(x^2+\frac{1}{x^2})}{(x^2+\frac{1}{x^2})}×(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^2-\frac{1}{x^2})(x^{16}+\frac{1}{x^{16}})(x^{4}+\frac{1}{x^{4}})$
$= \frac{1}{(x^2+\frac{1}{x^2})}×(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^{16}+\frac{1}{x^{16}})(x^{4}+\frac{1}{x^{4}})(x^{4}-\frac{1}{x^{4}})$
$= \frac{1}{(x^2+\frac{1}{x^2})}×(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^{16}+\frac{1}{x^{16}})(x^{8}-\frac{1}{x^{8}})$
$= \frac{1}{(x^2+\frac{1}{x^2})}×(x^{32}+\frac{1}{x^{32}})(x^{16}+\frac{1}{x^{16}})(x^{16}-\frac{1}{x^{16}})$
$= \frac{1}{(x^2+\frac{1}{x^2})}×(x^{32}+\frac{1}{x^{32}})(x^{32}-\frac{1}{x^{32}})$
$= \frac{1}{(x^2+\frac{1}{x^2})}×(x^{64}-\frac{1}{x^{64}})$
$= \frac{(x^{64}-\frac{1}{x^{64}})}{(x^2+\frac{1}{x^2})}$
Hence, the correct answer is $\frac{(x^{64}-\frac{1}{x^{64}})}{(x^2+\frac{1}{x^2})}$.