Question : If $x + \frac{1}{x} = \sqrt{3}$ , then the value of $x^{18} + x^{12} + x^{6} + 1$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3

Question

Question : If $x^{2} + \frac{1}{x^{2}} = \frac{7}{4}$ for $x > 0$ ; then what is the value of $x + \frac{1}{x}$ ?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$

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Answer 1

Correct Answer: 0

Solution : $x^{18} + x^{12} + x^{6} + 1$
= $x^{6} (x^{12} + 1) + 1 (x^{12} + 1)$
= $(x^{6} + 1) (x^{12} + 1)$ --------------------------(i)
$x + \frac{1}{x} = \sqrt{3}$
Squaring both sides, we get,
⇒ $(x + \frac{1}{x})^{2} = (\sqrt{3})^{2}$
⇒ $(x)^{2} + (\frac{1}{x})^{2} + 2 \times x \times \frac{1}{x} = 3$
⇒ $(x)^{2} + (\frac{1}{x})^{2} = 3 - 2 = 1$
Cubing both sides, we get,
$((x)^{2} + (\frac{1}{x})^{2})^{3} = 1^{3}$
⇒ $(x)^{6} + (\frac{1}{x})^{6} + 3 \times ((x)^{2} + (\frac{1}{x})^{2}) = 1^{3}$
⇒ $(x)^{6} + (\frac{1}{x})^{6} = 1 - 3$
⇒ $\frac{((x)^{12} + 1)}{(x)^{6}} = - 2$
⇒ $x^{12} + 1 + 2 x^{6} = 0$
⇒ $(x^{6} + 1)^{2} = 0$
⇒ $(x^{6} + 1) = 0$ -------------------------(ii)
Substituting (ii) in (i), we have,
$(x^{6} + 1) (x^{12} + 1)$ = 0
Thus, $x^{18} + x^{12} + x^{6} + 1$ = 0
Hence, the correct answer is 0.

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Question : If x+1x=3, then the value of x18+x12+x6+1 is:Option 1: 0Option 2: 1Option 3: 2Option 4: 3

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Question : If $x + \frac{1}{x} = \sqrt{3}$ , then the value of $x^{18} + x^{12} + x^{6} + 1$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3