1. **State the problem:** Simplify the expression $$\frac{14^{2} \cdot 7^{3} \cdot 98^{5} \cdot 18^{5} \cdot 63}{49^{3} \cdot 81^{4} \cdot 28^{3}}$$ 2. **Rewrite bases as prime factors:** - $14 = 2 \cdot 7$ - $7$ is prime - $98 = 14 \cdot 7 = (2 \cdot 7) \cdot 7 = 2 \cdot 7^{2}$ - $18 = 2 \cdot 3^{2}$ - $63 = 7 \cdot 9 = 7 \cdot 3^{2}$ - $49 = 7^{2}$ - $81 = 3^{4}$ - $28 = 4 \cdot 7 = 2^{2} \cdot 7$ 3. **Rewrite the entire expression with prime factors:** $$\frac{(2 \cdot 7)^{2} \cdot 7^{3} \cdot (2 \cdot 7^{2})^{5} \cdot (2 \cdot 3^{2})^{5} \cdot (7 \cdot 3^{2})}{(7^{2})^{3} \cdot (3^{4})^{4} \cdot (2^{2} \cdot 7)^{3}}$$ 4. **Apply exponents:** - $(2 \cdot 7)^{2} = 2^{2} \cdot 7^{2}$ - $(2 \cdot 7^{2})^{5} = 2^{5} \cdot 7^{10}$ - $(2 \cdot 3^{2})^{5} = 2^{5} \cdot 3^{10}$ - $(7^{2})^{3} = 7^{6}$ - $(3^{4})^{4} = 3^{16}$ - $(2^{2} \cdot 7)^{3} = 2^{6} \cdot 7^{3}$ 5. **Substitute back:** $$\frac{2^{2} \cdot 7^{2} \cdot 7^{3} \cdot 2^{5} \cdot 7^{10} \cdot 2^{5} \cdot 3^{10} \cdot 7 \cdot 3^{2}}{7^{6} \cdot 3^{16} \cdot 2^{6} \cdot 7^{3}}$$ 6. **Combine like bases in numerator:** - For base 2: $2^{2} \cdot 2^{5} \cdot 2^{5} = 2^{2+5+5} = 2^{12}$ - For base 7: $7^{2} \cdot 7^{3} \cdot 7^{10} \cdot 7^{1} = 7^{2+3+10+1} = 7^{16}$ - For base 3: $3^{10} \cdot 3^{2} = 3^{12}$ 7. **Combine like bases in denominator:** - For base 7: $7^{6} \cdot 7^{3} = 7^{9}$ - For base 3: $3^{16}$ - For base 2: $2^{6}$ 8. **Rewrite the fraction:** $$\frac{2^{12} \cdot 7^{16} \cdot 3^{12}}{7^{9} \cdot 3^{16} \cdot 2^{6}}$$ 9. **Subtract exponents for division:** - For base 2: $2^{12-6} = 2^{6}$ - For base 7: $7^{16-9} = 7^{7}$ - For base 3: $3^{12-16} = 3^{-4} = \frac{1}{3^{4}}$ 10. **Final simplified expression:** $$2^{6} \cdot 7^{7} \cdot \frac{1}{3^{4}} = \frac{2^{6} \cdot 7^{7}}{3^{4}}$$ 11. **Calculate numerical values:** - $2^{6} = 64$ - $7^{7} = 7^{3} \cdot 7^{4} = 343 \cdot 2401 = 823543$ - $3^{4} = 81$ 12. **Final answer:** $$\frac{64 \cdot 823543}{81} = \frac{52706752}{81}$$