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
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,
⇒ $(x+\frac{1}{x})^2 = (\sqrt{3})^2$
⇒ $(x)^2+(\frac{1}{x})^2+2×x×\frac{1}{x} = 3$
⇒ $(x)^2+(\frac{1}{x})^2 = 3-2 = 1$
Cubing both sides, we get,
$((x)^2+(\frac{1}{x})^2)^3 = 1^3$
⇒ $(x)^6+(\frac{1}{x})^6+3×((x)^2+(\frac{1}{x})^2) = 1^3$
⇒ $(x)^6+(\frac{1}{x})^6 = 1 - 3$
⇒ $\frac{((x)^{12}+1)}{(x)^6} = -2$
⇒ $x^{12}+1+2x^6 = 0$
⇒ $(x^{6}+1)^2 = 0$
⇒ $(x^{6}+1) = 0$ -------------------------(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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