1. **State the problem:** Simplify the expression $$\frac{3b^{-9}}{(3b^{-4}c^{-3})^{-3}}$$. 2. **Recall the rules:** - Negative exponents mean reciprocal: $$a^{-n} = \frac{1}{a^n}$$. - Power of a power: $$(a^m)^n = a^{mn}$$. - When dividing with the same base, subtract exponents: $$\frac{a^m}{a^n} = a^{m-n}$$. 3. **Simplify the denominator:** $$(3b^{-4}c^{-3})^{-3} = 3^{-3} (b^{-4})^{-3} (c^{-3})^{-3} = 3^{-3} b^{12} c^{9}$$. 4. **Rewrite the expression:** $$\frac{3b^{-9}}{3^{-3} b^{12} c^{9}} = 3^{1} b^{-9} \times 3^{3} b^{-12} c^{-9}$$ (since dividing by $3^{-3}$ is multiplying by $3^{3}$ and similarly for others). 5. **Combine like terms:** $$3^{1+3} b^{-9-12} c^{-9} = 3^{4} b^{-21} c^{-9}$$. 6. **Write with positive exponents:** $$3^{4} \times \frac{1}{b^{21} c^{9}} = \frac{81}{b^{21} c^{9}}$$. **Final answer:** $$\boxed{\frac{81}{b^{21} c^{9}}}$$