Given equation,

x² + 2x + 2 = 0

$$ \therefore $$  x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$

x = $$-$$ 1 $$ \pm $$ i

$$ \therefore $$  $$\alpha $$ = $$-$$ 1 + i

and $$\beta $$ = $$-$$ 1 $$-$$ i

Note :

x + iy = r (cos$$\theta $$ + isin$$\theta $$)

$$ \therefore $$  (x + iy)ⁿ = rⁿ (cosn$$\theta $$ + isinn$$\theta $$)

$$ \therefore $$  $$-$$ 1 + i = $$\sqrt 2 $$ [cos$${{3\pi } \over 4}$$ + isin$${{3\pi } \over 4}$$ ]

$$ \Rightarrow $$  ($$-$$ 1 + i)¹⁵ = $${\left( {\sqrt 2 } \right)^{15}}$$ [cos$$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$$]

And $$-$$1 $$-$$ i = $$\sqrt 2 $$ $$\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$$

= $$\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$$

$$ \therefore $$  ($$-$$1 $$-$$ i)¹⁵ = $${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$$

Now

$$\alpha $$¹⁵ + $$\beta $$¹⁵

= ($$-$$1 + i)¹⁵ + ($$-$$ 1 $$-$$ i)¹⁵

=  $${\left( {\sqrt 2 } \right)^{15}}$$ $$\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$$

= $$ - {\left( {\sqrt 2 } \right)^{14}}.2$$

= $$-$$ 2⁷ $$ \times $$ 2

= $$-$$ 2⁸

= $$-$$ 256