$\frac{{25}^3\ast {14}^2}{49\ast {10}^6}=\frac{(5^2{)}^3\ast (2\ast 7{)}^2}{7^2\ast (2\ast 5{)}^6}=\frac{5^6\ast 2^2\ast 7^2}{7^2\ast 2^6\ast 5^6}=\frac{1}{2^4}=\frac{1}{16}$

$\frac{{12}^2\ast {35}^3}{{28}^2\ast {15}^4}=\frac{(3\ast 4{)}^2\ast (5\ast 7{)}^3}{(4\ast 7{)}^2\ast (3\ast 5{)}^4}=\frac{3^2\ast 4^2\ast 5^3\ast 7^3}{4^2\ast 7^2\ast 3^4\ast 5^4}=\frac{7}{3^2\ast 5}=\frac{7}{9\ast 5}=\frac{7}{45}$

$\frac{{36}^3\ast {15}^2}{{18}^4\ast {10}^3}=\frac{(6^2{)}^3\ast (3\ast 5{)}^2}{(3\ast 6{)}^4\ast (2\ast 5{)}^3}=\frac{6^6\ast 3^2\ast 5^2}{3^4\ast 6^4\ast 2^3\ast 5^3}=\frac{6^2}{3^2\ast 2^3\ast 5}=\frac{(2\ast 3{)}^2}{3^2\ast 2^3\ast 5}=\frac{2^2\ast 3^2}{3^2\ast 2^3\ast 5}=\frac{1}{2\ast 5}=\frac{1}{10}$

$\frac{{22}^4\ast 3^3}{6^2\ast {121}^2}=\frac{(2\ast 11{)}^4\ast 3^3}{(2\ast 3{)}^2\ast ({11}^2{)}^2}=\frac{2^4\ast {11}^4\ast 3^3}{2^2\ast 3^2\ast {11}^4}=\frac{2^2\ast 3}{1}=4\ast 3=12$