Question

If $x+\frac{1}{x}=\sqrt{3}$, then the value of $x^{18}+x^{12}+x^6+1$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

$x+\frac{1}{x}=\sqrt{3}$
Cubing both the sides.
$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=3\sqrt{3}\Rightarrow x^3+\frac{1}{x^3}+3\left(\sqrt{3}\right)=3\sqrt{3}$

$x^3+\frac{1}{x^3}=0\Rightarrow \frac{x^6+1}{x^3}=0$
$x^6$ + 1 = 0

Now, $x^{18}+x^{12}+x^6+1$
$\Rightarrow x^{12}(x^6+1)+(x^6+1)$
From (1) $x^{12}$ (0) + (0) = 0