$$\left( {x + \frac{1}{x}} \right)$$ $$\left( {x - \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)$$  $$\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$$   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,
$$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$     $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$
$$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$