If $x^2-2 \sqrt{5} x+1=0$, then what is the value of $x^5+\frac{1}{x^5} ?$

$610 \sqrt{5}$

If $x^2-2 \sqrt{5} x+1=0$,

then what is the value of $x^5+\frac{1}{x^5}$ = ?

 x⁵ + $\frac{1}{x^5}$ = (x² + $\frac{1}{x^2}$) × (x³ + $\frac{1}{x^3}$) – (x + $\frac{1}{x}$)

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

If $x^2-2 \sqrt{5} x+1=0$,

then, divide equation by x on both sides,

x + \(\frac{1}{x}\) = $2\sqrt{5}$

 $x^3 +\frac{1}{x^3}$ = ( $2\sqrt{5}$)³ - 3 × $2\sqrt{5}$ = $34\sqrt{5}$

 $x^2 +\frac{1}{x^2}$ = ($2\sqrt{5}$)² – 2 = 18

 x⁵ + $\frac{1}{x^5}$ = 18 × $34\sqrt{5}$ - $2\sqrt{5}$

x⁵ + $\frac{1}{x^5}$ = $612\sqrt{5}$ - $2\sqrt{5}$ = $610 \sqrt{5}$