If $$x^4 + x^{-4} = 2599$$, then one of the values of $$x - x^{-1}$$, where x > 0, is equal to:

Given,

$$x^4+x^{-4}=2599$$

it can be written as

$$x^4+\dfrac{1}{x^4}=2599$$

adding 2 both sides

$$x^4+\dfrac{1}{x^4}+2=2599+2$$

or

$$x^4+\dfrac{1}{x^4}+2\times x^2\times (\dfrac{1}{x^2})=2601$$

$$(x^2+\dfrac{1}{x^2})^2=2601$$

$$x^2+\dfrac{1}{x^2}=\sqrt2601$$

$$x^2+\dfrac{1}{x^2}=51$$

now subtracting 2 both sides

$$x^2+\dfrac{1}{x^2}-2=51-2$$

or

$$x^2+\dfrac{1}{x^2}-2\times x^2\times \dfrac{1}{x^2}=49$$

$$(x-\dfrac{1}{x})^2=49$$

$$x-\dfrac{1}{x}=\sqrt49$$

$$x-\dfrac{1}{x}=7$$

or

$$x-x^{-1}=7$$