1. **State the problem:** Simplify the expression $$6v^2 \cdot \left(\frac{9u}{v}\right)^7 \cdot \left(\frac{v}{3u}\right)^{-2} \div \left(\frac{9u}{v}\right)^9$$. 2. **Recall the exponent rules:** - $\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$ - $x^{-n} = \frac{1}{x^n}$ - When dividing powers with the same base, subtract exponents: $x^a \div x^b = x^{a-b}$ - When multiplying powers with the same base, add exponents: $x^a \cdot x^b = x^{a+b}$ 3. **Rewrite each term with exponents:** $$6v^2 \cdot \frac{(9u)^7}{v^7} \cdot \left(\frac{v}{3u}\right)^{-2} \div \frac{(9u)^9}{v^9}$$ 4. **Simplify the negative exponent:** $$\left(\frac{v}{3u}\right)^{-2} = \left(\frac{3u}{v}\right)^2 = \frac{(3u)^2}{v^2} = \frac{9u^2}{v^2}$$ 5. **Substitute back:** $$6v^2 \cdot \frac{(9u)^7}{v^7} \cdot \frac{9u^2}{v^2} \div \frac{(9u)^9}{v^9}$$ 6. **Rewrite division as multiplication by reciprocal:** $$6v^2 \cdot \frac{(9u)^7}{v^7} \cdot \frac{9u^2}{v^2} \cdot \frac{v^9}{(9u)^9}$$ 7. **Group like terms:** $$6 \cdot 9 \cdot (9u)^7 \cdot u^2 \cdot v^{2 - 7 - 2 + 9} \cdot \frac{1}{(9u)^9}$$ Calculate the exponent of $v$: $$2 - 7 - 2 + 9 = 2$$ 8. **Simplify powers of 9 and u:** $$(9u)^7 \cdot u^2 = 9^7 u^7 \cdot u^2 = 9^7 u^{9}$$ Divide by $(9u)^9 = 9^9 u^9$: $$\frac{9^7 u^9}{9^9 u^9} = 9^{7-9} u^{9-9} = 9^{-2} = \frac{1}{9^2} = \frac{1}{81}$$ 9. **Combine constants:** $$6 \cdot 9 \cdot \frac{1}{81} = \frac{54}{81} = \frac{2}{3}$$ 10. **Combine all factors:** $$\frac{2}{3} \cdot v^2 = \frac{2}{3} v^2$$ **Final answer:** $$\boxed{\frac{2}{3} v^2}$$