If $$x + \frac{1}{x} = 2{\text{,}}$$ then the value of $$\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right)$$ is?
A. 20
B. 4
C. 8
D. 16
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = 2 \cr & {\text{Put x = 1}} \cr & \therefore {\text{1 + }}\frac{1}{{\left( 1 \right)}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisty}}} \right) \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = \left( {1 + 1} \right)\left( {1 + 1} \right) \cr & = 2 \times 2 \cr & = 4 \cr} $$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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