$$\left( {x + \frac{1}{x}} \right)$$ $$\left( {x - \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$$ is equal to ?
A. $${x^6} - \frac{1}{{{x^6}}}$$
B. $${x^8} - \frac{1}{{{x^8}}}$$
C. $${x^6} + \frac{1}{{{x^6}}}$$
D. $${x^8} + \frac{1}{{{x^8}}}$$
Answer: Option A
Solution (By Examveda Team)
Given expression,$$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$ $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$
$$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$
This Question Belongs to Arithmetic Ability >> Simplification
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