Why it's always x !!

Let $x = \cos \theta + i \sin \theta \Rightarrow \frac {1}{x} = \cos \theta - i \sin \theta$

Substitute them into the given equation yields $2 \cos \theta = \sqrt3 \Rightarrow \theta = \frac {\pi}{6}$

Apply De Moivre's Theorem: $x^6 = \cos (6 \theta) + i \sin (6 \theta) =-1$ , you should get the desired answer of $(-1)^3 + (-1)^2 + (-1)^1 + 1 = \boxed{0}$

$x+\dfrac {1}{x} = \sqrt{3}\quad \Rightarrow \left( x+\dfrac {1}{x} \right)^2 = \left( \sqrt{3} \right)^2 \quad \Rightarrow x^2 + 2 + \dfrac {1}{x^2} = 3$

$\Rightarrow x^2 + \dfrac {1}{x^2} = 1 \quad \Rightarrow x^2 \left( x^2+\dfrac {1}{x^2} \right) = x^2 \quad \Rightarrow x^4+1 = x^2$

$\Rightarrow x^2 \left( x^4+ 1 \right) = x^4 \quad \Rightarrow x^6 + x^2 = x^4 \quad \Rightarrow x^6 = x^4 - x^2 = -1$

$\Rightarrow x^{12} = ( x^6 )^2 = 1 \quad \Rightarrow x^{18} = (x^6)^3 = -1$

$\Rightarrow x^{18} + x^{12} + x^6 +1 = -1+1-1+1 = \boxed{0}$

Let The answer is a. Since x is not 0, we can write (x^18+x^12+x^6+1)/x^9 = a/x^9. (x^9 + 1/x^9) + (x^3+ 1/x^3) = a/x^9. Consider (x+1/x)^3 = 3.sqrt.(3), so (x^3 + 1/x^3) + 3(x+1/x) = 3.sqrt.(3), and we find (x^3 + 1/x^3) = 0, and consider (x^3 + 1/x^3)^3 = 0, or (x^9 + 1/x^9) + 3( (x^3 + 1/x^3) = 0, so (x^9 + 1/x^9) =0, we find a/x^9= 0, or a=0