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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?

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    $\begingroup$ I think this question has to be attempted without finding the value of the variable, unlike the linked question. Clearly, that would be cumbersome. There must be an easier way around this. $\endgroup$ – астон вілла олоф мэллбэрг Aug 2 '17 at 23:09
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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 $x^{18} = -1$.

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  • $\begingroup$ It's easier if you cube first rather than square, e.g. see my answer. $\endgroup$ – Gone Aug 2 '17 at 23:56
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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 $x^{18} = e^{\pm 3 \pi i} = -1$.

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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 $x^{18}=x^4-x^2=\cdots$

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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 $\,x^6 = -1\,$ $\Rightarrow\, x^{18} = -1.$

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$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 $x = \frac {\sqrt {3}}{2} \pm i\frac 12$

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