Unification

Factoring out $\large e^{i\dfrac{\pi}{4}}$ :

$\LARGE e^{i \frac {\pi} {4} } \left( e^{-i \frac {\pi} {8} }+e^{ i\frac{\pi} {8} } \right)$

Now this problem is purely computational. Using Euler's Formula, the question becomes

$\left(\cos{\left(\dfrac{\pi}{4}\right)}+ i \sin{\left(\dfrac{\pi}{4}\right)} \right) \left(\cos{\left(-\dfrac{\pi}{8}\right)} + \cos{\left(\dfrac{\pi}{8}\right)} + i \left( \sin{\left(-\dfrac{\pi}{8}\right)} + \sin{\left(\dfrac{\pi}{8}\right)} \right) \right)$

Using the facts that sine is an odd function and cosine is even function:

$\left(\cos{\left(\dfrac{\pi}{4}\right)}+ i \sin{\left(\dfrac{\pi}{4}\right)} \right) \left(2\cos{\left(\dfrac{\pi}{8}\right)} \right)$

Compute $\cos{\left(\dfrac{\pi}{8}\right)}$ using the $\cos{\left(\dfrac{x}{2}\right)} = \sqrt{\dfrac{1+\cos x}{2}}$ , an identity that can be derived from the power reducing formula. Evaluating $\cos{\left(\dfrac{\pi}{8}\right)}$ and $\cos{\left(\dfrac{\pi}{4}\right)}$

$\left(\sqrt{2+\sqrt{2}}\right)\left(\dfrac{1}{\sqrt 2} + i \dfrac{1}{\sqrt 2} \right)$

Factoring out $\dfrac{1}{\sqrt 2}$ :

$\left(\sqrt{\dfrac{2+\sqrt{2}}{2}}\right)\left(1 + i \right)$

This is the form asked for by the problem. Evaluating ${ \alpha }^{ a }\times { \beta }^{ b }\times \gamma$ , the answer is

$\boxed {8}$