1. **State the problem:** Find the limit $$\lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}.$$\n\n2. **Recall the formula and important rules:** The limit involves a cube root and a cubic polynomial in the denominator. Notice that $$x^3 + 8$$ can be factored as a sum of cubes: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ with $$a = x$$ and $$b = 2$$.\n\n3. **Factor the denominator:**\n$$x^3 + 8 = (x + 2)(x^2 - 2x + 4).$$\n\n4. **Evaluate the numerator and denominator at $$x = -2$$:**\n- Numerator: $$\sqrt[3]{-2 - 6} + 2 = \sqrt[3]{-8} + 2 = -2 + 2 = 0.$$\n- Denominator: $$(-2)^3 + 8 = -8 + 8 = 0.$$\n\nSince direct substitution gives $$\frac{0}{0}$$, we can apply algebraic manipulation or L'Hôpital's Rule.\n\n5. **Rewrite the numerator:** Let $$y = \sqrt[3]{x - 6}$$ so that $$y^3 = x - 6$$. Then the numerator is $$y + 2$$. When $$x \to -2$$, $$y \to \sqrt[3]{-8} = -2$$.\n\n6. **Use the identity for difference of cubes:**\n$$y^3 + 8 = (y + 2)(y^2 - 2y + 4).$$\nSince $$y^3 = x - 6$$, then\n$$y^3 + 8 = (x - 6) + 8 = x + 2.$$\n\n7. **Express numerator in terms of $$x$$:**\n$$y + 2 = \frac{y^3 + 8}{y^2 - 2y + 4} = \frac{x + 2}{y^2 - 2y + 4}.$$\n\n8. **Rewrite the original limit:**\n$$\lim_{x \to -2} \frac{y + 2}{x^3 + 8} = \lim_{x \to -2} \frac{\frac{x + 2}{y^2 - 2y + 4}}{(x + 2)(x^2 - 2x + 4)} = \lim_{x \to -2} \frac{1}{(y^2 - 2y + 4)(x^2 - 2x + 4)}.$$\n\n9. **Evaluate the limit by substituting $$x = -2$$ and $$y = -2$$:**\n- $$y^2 - 2y + 4 = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12.$$\n- $$x^2 - 2x + 4 = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12.$$\n\n10. **Calculate the limit:**\n$$\lim_{x \to -2} \frac{1}{12 \times 12} = \frac{1}{144}.$$\n\n**Final answer:** $$\boxed{\frac{1}{144}}.$$