1. **State the problem:** Simplify the expression $$\sqrt{9x^{8}y^{-4}} \times \sqrt[3]{8x^{6}y^{-3}}$$. 2. **Recall the formulas:** - The square root of a number is the same as raising it to the power of $\frac{1}{2}$: $$\sqrt{a} = a^{\frac{1}{2}}$$. - The cube root of a number is raising it to the power of $\frac{1}{3}$: $$\sqrt[3]{a} = a^{\frac{1}{3}}$$. 3. **Rewrite the expression using exponents:** $$\sqrt{9x^{8}y^{-4}} = (9x^{8}y^{-4})^{\frac{1}{2}}$$ $$\sqrt[3]{8x^{6}y^{-3}} = (8x^{6}y^{-3})^{\frac{1}{3}}$$ 4. **Apply the exponents to each factor inside the parentheses:** $$9^{\frac{1}{2}} \times (x^{8})^{\frac{1}{2}} \times (y^{-4})^{\frac{1}{2}} = 9^{\frac{1}{2}} x^{8 \times \frac{1}{2}} y^{-4 \times \frac{1}{2}} = 3 x^{4} y^{-2}$$ $$8^{\frac{1}{3}} \times (x^{6})^{\frac{1}{3}} \times (y^{-3})^{\frac{1}{3}} = 8^{\frac{1}{3}} x^{6 \times \frac{1}{3}} y^{-3 \times \frac{1}{3}} = 2 x^{2} y^{-1}$$ 5. **Multiply the two results:** $$ (3 x^{4} y^{-2}) \times (2 x^{2} y^{-1}) = 3 \times 2 \times x^{4+2} \times y^{-2 + (-1)} = 6 x^{6} y^{-3}$$ 6. **Rewrite with positive exponents where possible:** $$6 x^{6} \frac{1}{y^{3}} = \frac{6 x^{6}}{y^{3}}$$ **Final answer:** $$\boxed{\frac{6 x^{6}}{y^{3}}}$$