If x = √3 + √2 then the value of $$x^{3}-\frac{1}{x^{3}}$$ is

$$x = \sqrt{3} + \sqrt{2}$$

=> $$\frac{1}{x}$$ = $$\frac{1}{\sqrt{3}+\sqrt{2}}$$ = $$\frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2})}$$

=> $$\frac{1}{x}$$ = $$\frac{\sqrt{3} - \sqrt{2}}{3 - 2}$$ = $$\sqrt{3} - \sqrt{2}$$

Now, $$x - \frac{1}{x}$$ = $$\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}$$

= 2$$\sqrt{2}$$

Using, $$(x - \frac{1}{x})^{3}$$ = $$x^{3} - \frac{1}{x^{3}} - 3(x - \frac{1}{x})$$

=> $$x^{3}-\frac{1}{x^{3}}$$ = $$(2\sqrt{2})^{3} + 3(2\sqrt{2})$$

= $$16\sqrt{2} + 6\sqrt{2} = 22\sqrt{2}$$