Let $\alpha ,\beta$ be the roots of the equation  $x^2-\sqrt{2}x+2=0$. Then ${\alpha }^{14}-{\beta }^{14}$  is equal to 

Given,  $x^2-\sqrt{2}x+2=0$
 $$\begin{gathered} x=\frac{\sqrt{2}\pm \sqrt{2-8}}{2}=\frac{\sqrt{2}\pm \sqrt{6}i}{2}=\sqrt{2}(\frac{1}{2}\pm \frac{\sqrt{3}}{2}i) \\ \alpha =\sqrt{2}(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=\sqrt{2}{e}^{\frac{i\pi }{3}}\&\beta =\sqrt{2}{e}^{\frac{-i\pi }{3}} \\ {\alpha }^{14}=2^7{e}^{\frac{i14\pi }{3}}=128\left[e^{\frac{i2\pi }{3}}\right],{\beta }^{14}=128\left[e^{\frac{-i2\pi }{3}}\right] \\ {\alpha }^{14}-{\beta }^{14}=128\left(2i\right)\sin \left(\frac{2\pi }{3}\right)=128\sqrt{3}i \end{gathered}$$