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

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} + \frac{1}{x^{2}} = 1 \\ Then, {(x + \frac{1}{x})}^{2} = 1 + 2 \\ \Rightarrow x + \frac{1}{x} = \sqrt{3} \\ \Rightarrow x^{3} + \frac{1}{x^{3}} \\ = {(\sqrt{3})}^{3} - 3 \sqrt{3} \\ = 3 \sqrt{3} - 3 \sqrt{3} \\ = 0 \\ Then , x^{102} + x^{96} + x^{90} + x^{84} + x^{78} + x^{72} + 5 \\ = x^{96} (x^{6} + 1) + x^{84} (x^{6} + 1) + x^{78} (x^{6} + 1) + 5 \\ = 5 \end{aligned}

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