If $$x + \frac{1}{x} = 5{\text{,}}$$ then $${x^6}{\text{ + }}\frac{1}{{{x^6}}}$$ is?
A. 12098
B. 12048
C. 14062
D. 12092
Answer: Option A
Solution (By Examveda Team)
$$\eqalign{ & {\text{ }}x + \frac{1}{x} = 5 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( 5 \right)^3} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 110 \cr & \therefore {\text{Squaring both sides}} \cr & \Rightarrow {\left( {{x^3}{\text{ + }}\frac{1}{{{x^3}}}} \right)^2} = {\left( {110} \right)^2} \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} + 2 = 12100 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12100 - 2 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12098 \cr} $$Related Questions on Algebra
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$

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