1. **State the problem:** Simplify the expression $$\frac{(3a^{-2}b^3)^2 \times (3a^{-1} b^{-4})^{-1}}{(3a^2 b^{-2})^{-3}}$$. 2. **Recall the rules:** - Power of a power: $$(x^m)^n = x^{mn}$$ - Product of powers: $$x^m \times x^n = x^{m+n}$$ - Negative exponent: $$x^{-m} = \frac{1}{x^m}$$ - Quotient of powers: $$\frac{x^m}{x^n} = x^{m-n}$$ 3. **Simplify each part:** - First term: $$(3a^{-2}b^3)^2 = 3^2 a^{-4} b^6 = 9 a^{-4} b^6$$ - Second term: $$(3a^{-1} b^{-4})^{-1} = 3^{-1} a^{1} b^{4} = \frac{1}{3} a^{1} b^{4}$$ - Third term denominator: $$(3a^2 b^{-2})^{-3} = 3^{-3} a^{-6} b^{6} = \frac{1}{27} a^{-6} b^{6}$$ 4. **Multiply numerator terms:** $$9 a^{-4} b^{6} \times \frac{1}{3} a^{1} b^{4} = 3 a^{-4+1} b^{6+4} = 3 a^{-3} b^{10}$$ 5. **Divide numerator by denominator:** $$\frac{3 a^{-3} b^{10}}{\frac{1}{27} a^{-6} b^{6}} = 3 a^{-3} b^{10} \times 27 a^{6} b^{-6} = 81 a^{-3+6} b^{10-6} = 81 a^{3} b^{4}$$ 6. **Final answer:** $$\boxed{81 a^{3} b^{4}}$$