Let α, β be the roots of the equation x ^ 2 - 2x + 2 = 0 Then α ^...

Given,

${x}^{2} - \sqrt{2}x + 2 = 0$

$⇒ x = \dfrac{\sqrt{2} \pm \sqrt{6}i}{2}$

$⇒ x = \sqrt{2}\left(\dfrac{1 \pm \sqrt{3}i}{2}\right)$

$⇒ x = \sqrt{2}\left(\dfrac{1}{2} \pm \dfrac{\sqrt{3}i}{2}\right)$

$⇒ x = \sqrt{2}\left(cos\dfrac{\pi}{3} \pm isin\dfrac{\pi}{3}\right)$

So, $\alpha = \sqrt{2}\left(cos\dfrac{\pi}{3} + isin\dfrac{\pi}{3}\right) = \sqrt{2}e\dfrac{i \pi}{3}$

And $\beta = \sqrt{2}\left(cos\dfrac{\pi}{3} - isin\dfrac{\pi}{3}\right) = \sqrt{2}e\dfrac{- i \pi}{3}$

Now ${\alpha}^{14} + {\beta}^{14} = {\left(\sqrt{2}e\dfrac{i \pi}{3}\right)}^{14} + {\left(\sqrt{2}e\dfrac{- i \pi}{3}\right)}^{14}$

$⇒ {\alpha}^{14} + {\beta}^{14} = {\left(\sqrt{2}\right)}^{14}\left(e\dfrac{i14\pi}{3} + e\dfrac{- i14\pi}{3}\right)$

$⇒ {\alpha}^{14} + {\beta}^{14} = {\left(\sqrt{2}\right)}^{14}\left(cos\dfrac{14\pi}{3} + isin\dfrac{14\pi}{3} + cos\dfrac{14\pi}{3} - isin\dfrac{14\pi}{3}\right)$

$⇒ {\alpha}^{14} + {\beta}^{14} = {2}^{7}\left(2cos\dfrac{14\pi}{3}\right)$

$⇒ {\alpha}^{14} + {\beta}^{14} = {2}^{8}\left(\cos \left(4\pi + \dfrac{2\pi}{3}\right)\right)$

$⇒ {\alpha}^{14} + {\beta}^{14} = {2}^{8}\left(cos\dfrac{2\pi}{3}\right)$

$⇒ {\alpha}^{14} + {\beta}^{14} = - {2}^{8} \times \dfrac{1}{2} = - 128$