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 exponent rules:** - $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$ - $x^{-m} = \frac{1}{x^m}$ - When dividing powers with the same base, subtract exponents: $a^m \div a^n = a^{m-n}$ - When multiplying powers with the same base, add exponents: $a^m \cdot a^n = a^{m+n}$ 3. **Rewrite each term with exponents:** $$6v^2 \cdot \frac{(9u)^7}{v^7} \cdot \left( \frac{v}{3u} \right)^{-2} \cdot \frac{v^9}{(9u)^9}$$ 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:** $$6v^2 \cdot \frac{(9u)^7}{v^7} \cdot \frac{9u^2}{v^2} \cdot \frac{v^9}{(9u)^9}$$ 6. **Combine all factors:** $$6 \cdot v^2 \cdot \frac{(9u)^7}{v^7} \cdot \frac{9u^2}{v^2} \cdot \frac{v^9}{(9u)^9} = 6 \cdot 9 \cdot v^2 \cdot \frac{(9u)^7}{v^7} \cdot \frac{u^2}{v^2} \cdot \frac{v^9}{(9u)^9}$$ 7. **Group like terms:** - Coefficients: $6 \times 9 = 54$ - Powers of $v$: $v^2 \cdot v^{-7} \cdot v^{-2} \cdot v^9 = v^{2 - 7 - 2 + 9} = v^{2}$ - Powers of $u$: $u^7 \cdot u^2 \cdot u^{-9} = u^{7 + 2 - 9} = u^{0} = 1$ - Powers of 9: $9^7 \cdot 9^1 \cdot 9^{-9} = 9^{7 + 1 - 9} = 9^{-1} = \frac{1}{9}$ 8. **Combine all:** $$54 \cdot v^{2} \cdot 1 \cdot \frac{1}{9} = 6v^{2}$$ **Final answer:** $$6v^{2}$$