If $$x + \frac{1}{x} = \sqrt 3 {\text{,}}$$ then the value of x18 + x12 + x6 + 1 is?
A. 0
B. 1
C. 2
D. 3
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & \left( {{\text{Take cube on both sides}}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \therefore {x^6} = - 1 \cr & \therefore {x^{18}} + {x^{12}} + {x^6} + 1 \cr & = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 \cr & = - 1 + 1 - 1 + 1 \cr & = 0 \cr} $$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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