If $x +\frac{1}{x}=\sqrt{3}$, then the value of $x^{18} + x^{12} + x^6 + 1$ is:
Answer & explanation
Correct answer: option 1
If $x +\frac{1}{x}=\sqrt{3}$,
then the value of $x^{18} + x^{12} + x^6 + 1$
If $x +\frac{1}{x}=\sqrt{3}$,
then the value of x6 = -1 (always)
Put this value in required equation,
$x^{18} + x^{12} + x^6 + 1$ = (x6)3 + (x6)2 + x6 + 1 = -1 + 1 -1 + 1 = 0