Algebra or Geometry

Clearly easiest method is to put n=2 but here's a systematic approach.. $x+\dfrac{1}{x}=2\cos \frac{2π}{n}$ $\implies x^2-2\cos \frac{2π}{n}x+1=0$ $\implies (x-\cos \frac{2π}{n})^2=\cos^2 \frac{2π}{n}-1=-\sin^2 \frac{2π}{n}$ $\implies x=\color{#3D99F6}{\cos \frac{2π}{n}\pm i\sin \frac{2π}{n}=e^{\pm\frac{2\pi i}{n}}}$

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$\Large \therefore x^{n} + \frac{1}{x^{n}} =e^{\pm 2\pi i}+e^{\mp 2\pi i}$ $\Large =2=(2)\times n^0$ $\huge \therefore 2+0=\color{goldenrod}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{2}}}}}$

I prefer to approach this in terms of complex number. For $x = cos(y)+isin(y)$ , the $\frac{1}{x} = cos(y)-isin(y)$ which gives $x+\frac{1}{x} = 2cos(y)$

and

$x^n+\frac{1}{x^n} = 2cos(ny)$

As $y = \frac{2\pi}{n}$ , we will get the answer as $2cos(2\pi) =2$

We note that $\large cos \frac{2\pi}{n}$ = $\large \frac{x + x^{-1}}{2}$ . Hence, x = $\large e^\frac{2\pi}{n}$ .

Therefore the given expression $\large x^{n} + \frac{1}{x^{n}}$ always evaluates to 2.

We then deduce a to be 2 and b as zero for this to be possible for all possible n.

Similar solution with @Rishabh Cool 's

$\begin{aligned} x + \frac 1x & = 2\color{#3D99F6} \cos \frac {2\pi}n & \small \color{#3D99F6} \text{By Euler's formulla }e^{\theta i} = \cos \theta + i\sin \theta \\ & = 2\color{#3D99F6} \left(\frac {e^{\frac {2\pi}ni}+e^{-\frac {2\pi}ni}}2\right) \\ & = e^{\frac {2\pi}ni} + \frac 1{e^{\frac {2\pi}ni}} \\ \implies x & = e^{\frac {2\pi}ni} \end{aligned}$

Then we have:

$\begin{aligned} x^n + \frac 1{x^n} & = e^{\frac {2n\pi}ni}+e^{-\frac {2n\pi}ni} = e^{2\pi i} + e^{-2\pi i} = 2\cos 2\pi = 2 = 2n^0 \end{aligned}$

Therefore $a+b = 2+0 = \boxed{2}$ .