1. **Problem:** Simplify the expression $$\frac{6^3 \times 9^6 \times 3^4}{18^3}$$. 2. **Recall the rules:** - For powers with the same base: $$a^m \times a^n = a^{m+n}$$. - For division with the same base: $$\frac{a^m}{a^n} = a^{m-n}$$. - Express all numbers as powers of prime factors to simplify. 3. **Prime factorization:** - $$6 = 2 \times 3$$, so $$6^3 = (2 \times 3)^3 = 2^3 \times 3^3$$. - $$9 = 3^2$$, so $$9^6 = (3^2)^6 = 3^{12}$$. - $$3^4$$ is already prime base. - $$18 = 2 \times 3^2$$, so $$18^3 = (2 \times 3^2)^3 = 2^3 \times 3^{6}$$. 4. **Rewrite the expression:** $$\frac{6^3 \times 9^6 \times 3^4}{18^3} = \frac{2^3 \times 3^3 \times 3^{12} \times 3^4}{2^3 \times 3^6}$$ 5. **Combine powers of 3 in numerator:** $$3^{3 + 12 + 4} = 3^{19}$$ 6. **Simplify numerator and denominator:** $$\frac{2^3 \times 3^{19}}{2^3 \times 3^6} = 2^{3-3} \times 3^{19-6} = 2^0 \times 3^{13} = 3^{13}$$ 7. **Final answer:** $$\boxed{3^{13}}$$