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Algebra
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What is the simplified value of $$\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right){\text{is:}}$$

Answer & Solution
Correct Answer: Option D
$$\eqalign{ & \Rightarrow \left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right) \cr & \Rightarrow \frac{{\left[ {\left( {x - \frac{1}{x}} \right)\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^2} - \frac{1}{{{x^2}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^4} - \frac{1}{{{x^4}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^8} - \frac{1}{{{x^8}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^{16}} - \frac{1}{{{x^{16}}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^{32}} - \frac{1}{{{x^{32}}}}}}{{x - \frac{1}{x}}} \cr} $$
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