If $$x = \sqrt3 - \sqrt2$$, then the value of $$x^3 - x^{-3}$$ is:

Given $$x=\sqrt3 - \sqrt2$$ and we have to find $$x^3 - x^{-3}$$

As we know the identity, $$(x - \frac{1}{x})^3 = x^3 -( \frac{1}{x})^3 - 3(x)(\frac{1}{x})(x - \frac{1}{x}) $$

So we have to first find the value of $$\frac{1}{x}$$

$$\Rightarrow  \frac{1}{x} = \frac{1}{\sqrt3 - \sqrt2} = \frac{1\times (\sqrt3 + \sqrt2) }{(\sqrt3 - \sqrt2)\times(\sqrt3 + \sqrt2)} = \frac{\sqrt3+\sqrt2}{3-2} = \sqrt3 + \sqrt2$$  (Rationalizing the denominator)

$$\therefore x - \frac{1}{x} = (\sqrt3 - \sqrt2) - (\sqrt3 + \sqrt2) = -2\sqrt2$$

Now, using above identity

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

$$\Rightarrow (-2\sqrt2)^3 = x^3 -( \frac{1}{x})^3 - 3(-2\sqrt2) $$

$$\Rightarrow-16\sqrt2 = x^3 -( \frac{1}{x})^3 + 6\sqrt2$$

$$\therefore x^3 -( \frac{1}{x})^3 = -16\sqrt2-6\sqrt2=-22\sqrt2$$