If $${x^2} + \frac{1}{{{x^2}}} = 1{\text{,}}$$   then the value of $${x^{102}}$$ $$ + $$ $${x^{96}}$$ $$ + $$ $${x^{90}}$$ $$ + $$ $${x^{84}}$$ $$ + $$ $${x^{78}}$$ $$ + $$ $${x^{72}}$$ $$ + $$ $$5$$ is?

Solution(By Examveda Team)

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 1 \cr & {\text{Then, }}{\left( {x + \frac{1}{x}} \right)^2} = 1 + 2 \cr & \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\sqrt 3 } \right)^3} - 3\sqrt 3 \cr & = 3\sqrt 3 - 3\sqrt 3 \cr & = 0 \cr & {\text{Then,}} \cr & {x^{102}} + {x^{96}} + {x^{90}} + {x^{84}} + {x^{78}} + {x^{72}} + 5 \cr} $$
  $$ = {x^{96}}\left( {{x^6} + 1} \right) + $$    $${x^{84}}\left( {{x^6} + 1} \right) + $$   $${x^{72}}\left( {{x^6} + 1} \right) + $$   $$5$$
  $$ = 5$$