A little weirder!

I have another easy solution.

Then,

That's the solution!!

$\begin{aligned} x + \frac{1}{x} & = \sqrt{3} \\ \Rightarrow \left( x + \frac{1}{x} \right)^3 & = \left(\sqrt{3}\right)^3 \\ x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} & = 3\sqrt{3} \\ x^3 + 3\sqrt{3} + \frac{1}{x^3} & = 3\sqrt{3} \\ x^3 + \frac{1}{x^3} & = 0 \end{aligned}$

And:

$\begin{aligned} \left( x^3 + \frac{1}{x^3} \right)^{3} & = 0 \\ x^9 + 3x^3 + \frac{3}{x^3} + \frac{1}{x^9} & = 0 \\ x^9 + 3(0) + \frac{1}{x^9} & = 0 \\ \Rightarrow x^9 + \frac{1}{x^9} & = 0 \end{aligned}$

Similarly,

$\begin{aligned} \left( x^3 + \frac{1}{x^3} \right)^{5} & = 0 \\ x^{15} + 5x^9 + 10x^3 + \frac{10}{x^3} + \frac{5}{x^9} + \frac{1}{x^{15}} & = 0 \\ x^{15} + \frac{1}{x^{15}} + 5\left(x^9 + \frac{1}{x^9}\right) + 10 \left(x^3 + \frac{1}{x^3}\right) & = 0 \\ x^{15} + \frac{1}{x^{15}} + 5\left(0\right) + 10 \left(0 \right) & = 0 \\ \Rightarrow x^{15} + \frac{1}{x^{15}} & = \boxed{0} \end{aligned}$

Let $x=\cos(\theta)+i\sin(\theta)$ . Then $x+\frac{1}{x}=2\cos(\theta)=\sqrt{3}$ . Take $\theta=\frac{\pi}{6}$ . Now $x^{15}=\cos(15\theta)+i\sin(15\theta)=0+i$ . Then $x^{15}+\frac{1}{x^{15}}=2\cos(15\theta)=0$ .

I used De Moivres Formula here.. Solving for the positive root we get, sqrt3/2 +(1/2)(i) then used The formula for the 15th power

We get the solution as $iw,iw^2$ put in the equation, use $w^3=1$ as $w$ is cube root of unity, $i^3=-i$ to get result as $0$ .

Assuming that $f(n) = x^{n} + y^{n}$ , so we have: $1. f(1)=\sqrt{3}. \quad$ $2. f(2)=1. \quad$ $3. f(n+1)=\sqrt{3}f(n) - f(n-1). \quad$ By resolving this we can see that $f(n+6) = f(n)$ with every $n>=3$ , so $f(15)=f(3)=0.$

$(x+\frac { 1 }{ x } )=\sqrt { 3 }$

${ (x+\frac { 1 }{ x } ) }^{ 3 }={ x }^{ 3 }+3(x+\frac { 1 }{ x } )+\frac { 1 }{ { x }^{ 3 } } \\ { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } -3(x+\frac { 1 }{ x } )\\ \qquad \qquad =3\sqrt { 3 } -3\sqrt { 3 } \\ \qquad \qquad =0$

It can also be observed that ${ x }^{ 15 }+\frac { 1 }{ { x }^{ 15 } } =({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } )({ x }^{ 12 }+\frac { 1 }{ { x }^{ 12 } } )\quad -\quad ({ x }^{ 9 }+\frac { 1 }{ { x }^{ 9 } } )$

$({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } )({ x }^{ 12 }+\frac { 1 }{ { x }^{ 12 } } )\quad =\quad 0\\ \because \quad ({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } )=0$

${ ({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ) }^{ 3 }={ x }^{ 9 }+\frac { 1 }{ { x }^{ 9 } } +3({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } )\\ { x }^{ 9 }+\frac { 1 }{ { x }^{ 9 } } =3\sqrt { 3 } -3\sqrt { 3 } \\ \qquad \qquad =0$

Thus

${ x }^{ 15 }+\frac { 1 }{ { x }^{ 15 } } =0-0\\ \qquad \qquad \quad =0$

Plus it can also be generalised that when

${ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =\sqrt { 3 }$

${ x }^{ y }+\frac { 1 }{ { x }^{ y } } =0\\ y={ 3 }^{ n }\quad$

when n is element of integer and $n\neq 0$

Clearly, $x\neq 0$ . Multiplying both sides by $x$ then using quadratics equation on $x^2-\sqrt{3}x+1=0$ , we have $x=\dfrac{\sqrt{3}\pm i}{2}=cis \left(\pm \dfrac{\pi}{6}\right)$ .

It doesn't matter which root we use for $x$ because $r_1^{15}=\dfrac{1}{r_2^{15}}$ if we let those two roots be $r_1$ and $r_2$ . So WLOG, let $x=cis\left(\dfrac{\pi}{6}\right)$ . Then $x^{15}+\dfrac{1}{x^{15}}=cis\left(\dfrac{15\pi}{6}\right)+cis\left(-\dfrac{15\pi}{6}\right)=2\cos\left(\dfrac{5\pi}{2}\right)=0$ .

nice problem