If $$x^2 - 4x + 1 = 0$$, then what is the value of $$(x^6 + x^{-6})$$?

 Given $$x^2 - 4x +1 = 0 $$ the $$(x^6 +x^{-6})$$ = ?

$$ x^2 - 4x +1 =0$$

$$\Rightarrow x^2 - 4x = -1$$

$$\Rightarrow x(x-4) = -1$$

$$\Rightarrow x-4 = \dfrac {-1}{x} $$

$$\Rightarrow x+\dfrac{1}{x}= 4 $$ '...........  (1)

Taking cube of both side of the equestion (1)

$$(x+(\dfrac{1} {x}))^3 = (4)^3 $$

$$\Rightarrow x^3 +( \dfrac {1}{x})^3 + 3x \dfrac {1}{x} (x+\dfrac{1}{x}) = 64 $$

$$\Rightarrow x^3 +( \dfrac{1}{x})^3+ 3 (4) = 64$$

$$\Rightarrow x^3 + (\dfrac{1}{x})^3 = 64-12$$

$$\Rightarrow x^3 + (\dfrac{1}{x})^3 = 52$$   ...........  (2)

on squaring both side equestion (2) both side 

$$ (x^3+\dfrac {1}{x}^3)^2 = (52)^2 $$

$$\Rightarrow x^6 +(\dfrac{1}{x})^6+2 = 2704$$

$$\Rightarrow x^6+(\dfrac{1}{x})^6 = 2704 -2 $$

$$\Rightarrow x^6+(\dfrac {1}{x})^6 = 2702 $$Ans