If $x + \frac{1}{x} = \sqrt{3}$, then find $x^{18}$

Question

If $x + \frac{1}{x} = \sqrt{3}$, then find $x^{18}$. This is sorta like If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$.
So my question is, do I solve my question the same way as the question in the link?

Answer 1

As I thought, it will get easier on squaring both sides, which gives: $$ x^2 + \frac 1{x^2} = 3 -2 = 1 $$ Which then on cubing gives : $$ x^6 + \frac 1{x^6} + 3\left(x^2 + \frac 1{x^2}\right)= 1 \implies x^6 + \frac 1{x^6} = -2 $$ And then on cubing again gives : $$ x^{18} + \frac 1{x^{18}} + 3\left(x^6 + \frac 1{x^6}\right) = -8 \implies x^{18} + \frac 1{x^{18}} = -2 $$

By inspection above , or the quadratic formula, it's easy to see that $x^{18} = -1$.

Answer 2

You can just multiply by $x$ and use the quadratic formula. $x^2 + 1 = \sqrt{3}x \implies x^2 - \sqrt{3}x + 1 = 0$.

Then $x = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm i}{2}$. Luckily, we know that this is $e^{\frac{\pm i\pi}{6}}$, so we have $x^{18} = e^{\pm 3 \pi i} = -1$.

Answer 3

Hint:  squaring the equality gives:

$$x^2+\frac{1}{x^2}+2=3 \;\;\implies\;\; x^4 = x^2-1$$

Then $x^8=(x^2-1)^2=x^4-2x^2+1=-x^2\,$, $x^{16}=x^4=x^2-1\,$, and $x^{18}=x^4-x^2=\cdots$

Answer 4

With $\,y = x^{-1}$, $\,x\!+\!y = \sqrt3,\, xy = 1\,$ so $\,x^3\!+\!y^3 = (x\!+\!y)^3\! - 3xy(x\!+\!y) = (\sqrt 3)^3\!-3(1)\sqrt 3 = 0$ thus $\,x^3 = -y^3 = -1/x^3$ so $\,x^6 = -1\,$ $\Rightarrow\, x^{18} = -1.$

Answer 5

$x + \frac 1x \ge 2 $ for all real $x.$

If $x\in \mathbb C$

Squaring both sides:

$x^2 + \frac 1{x^2} + 2 = 3\\ x^4 - x^2 + 1 = 0\\ x^2 = \frac 12 + i \frac {\sqrt 3}{2}\\ x^2 = e^{\frac {\pi}{3}i} \\ x^{18} = e^{3\pi i} = -1$

Can ignore the $\pm$ in the application of the quadratic formula above as one will apply to $x^2$ and the other to $\frac 1{x^2}$

And if you care $x = \frac {\sqrt {3}}{2} \pm i\frac 12$