If $${x^2} - x + 1 = 0$$    then the value of $${\sum\limits_{n = 1}^5 {\left( {{x^n} + \frac{1}{{{x^n}}}} \right)} ^2}$$   is

Solution :
$$\eqalign{ & x = \frac{{1 \pm \sqrt {3i} }}{2} = \cos \frac{\pi }{3} \pm i\sin \frac{\pi }{3} \cr & \therefore \,\,{x^{2n}} = \cos \frac{{2n\pi }}{3} \pm i\sin \frac{{2n\pi }}{3} \cr & {\text{or,}}\,\,{\left( {{x^n} + \frac{1}{{{x^n}}}} \right)^2} = {x^{2n}} + {x^{ - 2n}} + 2 \cr & {\text{or,}}\,\,{\left( {{x^n} + \frac{1}{{{x^n}}}} \right)^2} = \cos \frac{{2n\pi }}{3} \pm i\sin \frac{{2n\pi }}{3} + \cos \frac{{2n\pi }}{3} \mp i\sin \frac{{2n\pi }}{3} + 2 \cr & {\text{or,}}\,\,{\left( {{x^n} + \frac{1}{{{x^n}}}} \right)^2} = 2 + 2\cos \frac{{2n\pi }}{3}. \cr} $$
∴ the value $$ = 10 + 2\left\{ {\cos \frac{{2\pi }}{3} + \cos \frac{{4\pi }}{3} + \cos \frac{{6\pi }}{3} + \cos \frac{{8\pi }}{3} + \cos \frac{{10\pi }}{3}} \right\}.$$