If $$x^2 - 3x + 1 = 0$$, then what is the value of $$x^6 + \frac{1}{x^6}$$?

$$x^2 - 3x + 1 = 0$$

Divide both sides by x,

$$x - 3 + \frac{1}{x} = 0$$

$$x + \frac{1}{x} = 3$$

On taking both sides cube,

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

$$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = 27$$

$$(\because (a + b)^3 = a^3 + b^3 + 3ab(a + b))$$

$$x^3 + \frac{1}{x^3} + 3(3) = 27$$

$$x^3 + \frac{1}{x^3} = 18$$

on taking square both sides,

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

$$x^6 + \frac{1}{x^6} + 2 = 324$$

$$(\because (a + b)^2 = a^2 + b^2 + 2ab)$$

$$x^6 + \frac{1}{x^6} = 322$$