1. **State the problem:** Simplify the expression $$\frac{(8x^6y^{-3})^{\frac{1}{3}}}{\sqrt{16x^8}}$$. 2. **Recall the rules:** - Power of a product: $ (ab)^m = a^m b^m $ - Power of a power: $ (a^m)^n = a^{mn} $ - Negative exponent: $ a^{-m} = \frac{1}{a^m} $ - Square root as exponent: $ \sqrt{a} = a^{\frac{1}{2}} $ 3. **Simplify numerator:** $$ (8x^6y^{-3})^{\frac{1}{3}} = 8^{\frac{1}{3}} (x^6)^{\frac{1}{3}} (y^{-3})^{\frac{1}{3}} $$ $$ = 2 \cdot x^{6 \cdot \frac{1}{3}} \cdot y^{-3 \cdot \frac{1}{3}} = 2x^2 y^{-1} $$ 4. **Simplify denominator:** $$ \sqrt{16x^8} = (16x^8)^{\frac{1}{2}} = 16^{\frac{1}{2}} (x^8)^{\frac{1}{2}} = 4x^4 $$ 5. **Rewrite the expression:** $$ \frac{2x^2 y^{-1}}{4x^4} $$ 6. **Simplify the fraction:** $$ = \frac{\cancel{2} x^2 y^{-1}}{\cancel{4} x^4} = \frac{1}{2} \cdot \frac{x^2}{x^4} \cdot y^{-1} $$ 7. **Simplify powers of $x$:** $$ \frac{x^2}{x^4} = x^{2-4} = x^{-2} $$ 8. **Combine all:** $$ \frac{1}{2} x^{-2} y^{-1} = \frac{1}{2} \cdot \frac{1}{x^2} \cdot \frac{1}{y} = \frac{1}{2x^2 y} $$ **Final answer:** $$\frac{1}{2x^2 y}$$