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