If $$1 + {x^2} = \sqrt {3x} $$    then$$\sum\limits_{n = 1}^{24} {{{\left( {{x^n} - \frac{1}{{{x^n}}}} \right)}^2}} $$    is equal to

Solution :
$$\eqalign{ & 1 - \sqrt 3 x + {x^2} = 0 \cr & x = \frac{{\sqrt 3 \pm i}}{2} \cr & x = \frac{{\sqrt 3 + i}}{2} = - i\overline w {\text{ and }}\overline x = \frac{{\sqrt 3 - i}}{2} = iw \cr & {\text{Now, }}{\left( {{x^n} - \frac{1}{{{x^n}}}} \right)^2} \cr & = {x^{2n}} + {\overline w ^{2n}} - 2 \cr & = - {\overline w ^{2n}} - {w^{2n}} - 2 \cr & = - 2 - \left[ {{{\overline w }^{2n}} + {w^{2n}}} \right] \cr & {\text{Now, }}\overline w = {w^2}{\text{ and }}1 + w + {w^2} = 0 \cr & {\text{Hence,}} \cr & - 2 - \left[ {{{\overline w }^{2n}} + {w^{2n}}} \right] \cr & = - 2 - \left[ {{w^{4n}} + {w^{2n}}} \right] \cr & = - 2 - \left[ { - 1} \right] \cr & = - 2 + 1 \cr & = - 1 \cr & {\text{Hence, }}\sum\limits_{n = 1}^{24} {{{\left( {{x^n} - \frac{1}{{{x^n}}}} \right)}^2}} = - 24 \cr} $$