1. **Problem:** Find the limit $$\lim_{x \to a} \frac{x^3 - a^3}{a^6 - x^6}$$ where $a \neq 0$. 2. **Formula and rules:** Recall the difference of cubes factorization: $$x^3 - a^3 = (x - a)(x^2 + ax + a^2)$$ and the difference of sixth powers factorization: $$a^6 - x^6 = (a^3 - x^3)(a^3 + x^3)$$ which can be further factored using difference and sum of cubes. 3. **Factor numerator and denominator:** $$\frac{x^3 - a^3}{a^6 - x^6} = \frac{(x - a)(x^2 + ax + a^2)}{(a^3 - x^3)(a^3 + x^3)}$$ Note that: $$a^3 - x^3 = -(x^3 - a^3) = -(x - a)(x^2 + ax + a^2)$$ 4. **Rewrite denominator:** $$a^6 - x^6 = (a^3 - x^3)(a^3 + x^3) = -(x - a)(x^2 + ax + a^2)(a^3 + x^3)$$ 5. **Simplify the fraction:** $$\frac{(x - a)(x^2 + ax + a^2)}{-(x - a)(x^2 + ax + a^2)(a^3 + x^3)} = \frac{\cancel{(x - a)}\cancel{(x^2 + ax + a^2)}}{-\cancel{(x - a)}\cancel{(x^2 + ax + a^2)}(a^3 + x^3)} = \frac{1}{-(a^3 + x^3)}$$ 6. **Evaluate the limit as $x \to a$:** $$\lim_{x \to a} \frac{1}{-(a^3 + x^3)} = \frac{1}{-(a^3 + a^3)} = \frac{1}{-2a^3} = -\frac{1}{2a^3}$$ **Final answer:** $$\boxed{-\frac{1}{2a^3}}$$ This corresponds to option (C).