If $$\frac{8x}{2x^2 + 7x - 2} = 1, x > 0$$, then what is the value of $$x^3 + \frac{1}{x^3}$$?

Given that $$\frac{8x}{2x^2 + 7x - 2} = 1$$

$$\Rightarrow 8x= 2 x^2+ 7x-2 $$

$$\Rightarrow 2x^2-x-2 = 0 $$

$$\Rightarrow x - \dfrac{1}{2} - \dfrac{1}{x} = 0 $$

$$\Rightarrow x - \dfrac{1}{x}= \dfrac{1}{2} $$

then $$ x +\dfrac {1}{x} = \sqrt {(x-\dfrac{1}{x})^2 +4x\dfrac{1}{x}} $$

$$\Rightarrow \sqrt{\dfrac{1}{4} + 4} $$

$$\Rightarrow \dfrac{\sqrt{17}}{2}$$

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

$$\Rightarrow (\dfrac{\sqrt{17}}{2})^3 - 3 \dfrac{\sqrt 17}{2} $$

$$\Rightarrow 17 \dfrac{\sqrt{17}} {8}- 3 \dfrac {\sqrt {17}}{2}$$

$$\Rightarrow \dfrac{5}{8} \sqrt{17} $$ Ans