1. **State the problem:** We need to find the value of the expression $$x^3 - 3x$$ where $$x = \sqrt[3]{2 - \sqrt{3}} - \sqrt[3]{2 + \sqrt{3}}$$. 2. **Recall the formula:** For any real numbers $$a$$ and $$b$$, the identity $$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$$ holds. 3. **Set:** $$a = \sqrt[3]{2 - \sqrt{3}}$$ and $$b = \sqrt[3]{2 + \sqrt{3}}$$, so $$x = a - b$$. 4. **Calculate $$a^3$$ and $$b^3$$:** $$a^3 = 2 - \sqrt{3}$$ $$b^3 = 2 + \sqrt{3}$$ 5. **Calculate $$a^3 - b^3$$:** $$a^3 - b^3 = (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}$$ 6. **Calculate $$ab$$:** $$ab = \sqrt[3]{(2 - \sqrt{3})(2 + \sqrt{3})} = \sqrt[3]{4 - (\sqrt{3})^2} = \sqrt[3]{4 - 3} = \sqrt[3]{1} = 1$$ 7. **Use the identity:** $$x^3 = a^3 - b^3 - 3ab(a - b) = -2\sqrt{3} - 3 \cdot 1 \cdot x = -2\sqrt{3} - 3x$$ 8. **Rearrange to find $$x^3 - 3x$$:** $$x^3 - 3x = -2\sqrt{3}$$ **Final answer:** $$\boxed{-2\sqrt{3}}$$