1. **Problem statement:** Find the limit $$\lim_{x \to 0} \frac{x}{\sqrt[3]{x+8} + \sqrt[3]{x-8}}.$$\n\n2. **Recall the formula for cube roots:** For any real number $a$, $\sqrt[3]{a}$ is the cube root of $a$. Also, note that $8 = 2^3$.\n\n3. **Evaluate the cube roots at $x=0$:**\n$$\sqrt[3]{0+8} = \sqrt[3]{8} = 2,$$\n$$\sqrt[3]{0-8} = \sqrt[3]{-8} = -2.$$\n\n4. **Substitute $x=0$ into the denominator:**\n$$\sqrt[3]{0+8} + \sqrt[3]{0-8} = 2 + (-2) = 0.$$\n\nSince direct substitution gives a denominator of zero, the limit is an indeterminate form $\frac{0}{0}$, so we need to simplify.\n\n5. **Use the identity for difference of cubes:**\nRecall that for any $a$ and $b$,\n$$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$\n\nLet\n$$a = \sqrt[3]{x+8}, \quad b = \sqrt[3]{x-8}.$$\nThen\n$$a^3 = x+8, \quad b^3 = x-8,$$\nso\n$$a^3 - b^3 = (x+8) - (x-8) = 16.$$\n\n6. **Express $a - b$ in terms of $a^3 - b^3$:**\n$$a^3 - b^3 = (a - b)(a^2 + ab + b^2) \implies a - b = \frac{16}{a^2 + ab + b^2}.$$\n\n7. **We want $a + b$, but we have $a - b$. Let's find $a + b$ by considering the original denominator:**\nThe denominator is $a + b$. We can write\n$$a + b = ?$$\n\n8. **Use the fact that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$:**\nCalculate $a^3 + b^3$:\n$$a^3 + b^3 = (x+8) + (x-8) = 2x.$$\n\nSo,\n$$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 2x.$$\n\n9. **We want to find $a + b$, so:**\n$$a + b = \frac{2x}{a^2 - ab + b^2}.$$\n\n10. **Rewrite the original limit:**\n$$\lim_{x \to 0} \frac{x}{a + b} = \lim_{x \to 0} \frac{x}{\frac{2x}{a^2 - ab + b^2}} = \lim_{x \to 0} \frac{x \cdot (a^2 - ab + b^2)}{2x}.$$\n\n11. **Cancel $x$ in numerator and denominator:**\n$$\lim_{x \to 0} \frac{\cancel{x} \cdot (a^2 - ab + b^2)}{2 \cancel{x}} = \lim_{x \to 0} \frac{a^2 - ab + b^2}{2}.$$\n\n12. **Evaluate $a$ and $b$ at $x=0$:**\n$$a = \sqrt[3]{8} = 2, \quad b = \sqrt[3]{-8} = -2.$$\n\nCalculate each term:\n$$a^2 = 2^2 = 4,$$\n$$ab = 2 \times (-2) = -4,$$\n$$b^2 = (-2)^2 = 4.$$\n\nSo,\n$$a^2 - ab + b^2 = 4 - (-4) + 4 = 4 + 4 + 4 = 12.$$\n\n13. **Final limit value:**\n$$\lim_{x \to 0} \frac{x}{\sqrt[3]{x+8} + \sqrt[3]{x-8}} = \frac{12}{2} = 6.$$\n\n**Answer:** $6$