Given that $$x^8 - 34x^4 + 1 = 0, x > 0$$. What is the value of $$(x^3 - x^{-3})$$?

$$x^8-34x^4+1=0$$

$$x^8+1=34x^4$$

$$x^4+\frac{1}{x^4}=34$$

$$x^4+\frac{1}{x^4}+2=36$$

$$\left(x^2+\frac{1}{x^2}\right)^2=36$$

$$x^2+\frac{1}{x^2}=6$$

$$x^2+\frac{1}{x^2}-2=4$$

$$\left(x-\frac{1}{x}\right)^2=4$$

$$x-\frac{1}{x}=2$$........(1)

$$\left(x-\frac{1}{x}\right)^3=8$$

$$x^3-\frac{1}{x^3}-3.x.\frac{1}{x}\left(x-\frac{1}{x}\right)=8$$

$$x^3-\frac{1}{x^3}-3\left(2\right)=8$$

$$x^3-\frac{1}{x^3}-6=8$$

$$x^3-\frac{1}{x^3}=14$$

Hence, the correct answer is Option A