Let $$\alpha \,{\text{and }}\beta $$  be two roots of the equation $${x^2} + 2x + 2 = 0,$$    then $${\alpha ^{15}} + {\beta ^{15}}$$  is equal to:

Solution :
Consider the equation
$$\eqalign{ & {x^2} + 2x + 2 = 0 \cr & x = {\left( {\sqrt 2 {e^{i\frac{{3\pi }}{4}}}} \right)^{15}} + {\left( {\sqrt 2 {e^{ - i\frac{{3\pi }}{4}}}} \right)^{15}} \cr & {\text{Let }}\alpha = - 1 + i,\,\,\beta = - 1 - i \cr & {\alpha ^{15}} + {\beta ^{15}} = {\left( { - 1 + i} \right)^{15}} + {\left( { - 1 - i} \right)^{15}} \cr & = {\left( {\sqrt 2 {e^{i\frac{{3\pi }}{4}}}} \right)^{15}} + {\left( {\sqrt 2 {e^{ - i\frac{{3\pi }}{4}}}} \right)^{15}} \cr & = {\left( {\sqrt 2 } \right)^{15}}\left[ {{e^{\frac{{i45\pi }}{4}}} + {e^{\frac{{ - i45\pi }}{4}}}} \right] \cr & = {\left( {\sqrt 2 } \right)^{15}}.2\cos \frac{{45\pi }}{4} \cr & = {\left( {\sqrt 2 } \right)^{15}}.2\cos \frac{{3\pi }}{4} \cr & = \frac{{ - 2}}{{\sqrt 2 }}{\left( {\sqrt 2 } \right)^{15}} \cr & = - 2{\left( {\sqrt 2 } \right)^{14}} \cr & = - 256 \cr} $$