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

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

Divide by x,

$$x - 2\sqrt{5} + \frac{1}{x} = 0$$

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

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

$$(x + \frac{1}{x})^2 = 20$$ 

$$x^2 + (\frac{1}{x})^2 + 2 = 20$$ 

$$x^2 + (\frac{1}{x})^2 = 18$$ ----(2)

From eq(1),

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

$$x^3 + (\frac{1}{x})^3 + 3(x + \frac{1}{x}) = 40\sqrt{5}$$

$$x^3 + (\frac{1}{x})^3 = 40\sqrt{5} - 3(2\sqrt{5}) = 34\sqrt{5}$$ ---(3)

From eq(2) and (3),

$$(x^2 + (\frac{1}{x})^2)(x^3 + (\frac{1}{x})^3) = (18)(34\sqrt{5})$$

$$(x^5 + \frac{1}{x} + x + (\frac{1}{x})^5) = 612\sqrt{5}$$

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