If $$x + \frac{1}{x} = 3{\text{,}}$$ then the value of $$\left( {{x^5} + \frac{1}{{{x^5}}}} \right)\,{\text{is?}}$$
A. 322
B. 126
C. 123
D. 113
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \left( {{\text{Squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{On cubing both sides}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 18 \times 7 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + \left( {x + \frac{1}{x}} \right) = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + 3 = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) = 123 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
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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