Since, $\alpha, \beta$ are the roots of the equation
$$
\begin{aligned}
x^2-4 x+8 & =0 \\
\therefore \quad \alpha, \beta & =\frac{4 \pm \sqrt{16-32}}{2}=\frac{4 \pm 4 i}{2}=2 \pm 2 i \\
\Rightarrow \quad \alpha, \beta & =2 \sqrt{2}\left(\frac{1}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}}\right) \\
& =2 \sqrt{2}\left(\cos \frac{\pi}{4} \pm i \sin \frac{\pi}{4}\right) \\
\text { Now, } \alpha^{2 n} & =(2 \sqrt{2})^{2 n}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{2 n}, n \in N \\
& =(2 \sqrt{2})^{2 n}\left(\cos \frac{n \pi}{2}+i \sin \frac{n \pi}{2}\right)
\end{aligned}
$$

Similarly, $\beta^{2 n}=(2 \sqrt{2})^{2 n}\left(\cos \frac{n \pi}{2}-i \sin \frac{n \pi}{2}\right)$
$$
\begin{aligned}
\therefore \quad \alpha^{2 n}+\beta^{2 n} & =(2 \sqrt{2})^{2 n} \cdot 2 \cos \frac{n \pi}{2} \\
& =\left(2^{3 / 2}\right)^{2 n} \cdot 2 \cos \frac{n \pi}{2} \\
\Rightarrow \quad \alpha^{2 n}+\beta^{2 n} & =2^{3 n+1} \cos \frac{n \pi}{2}
\end{aligned}
$$