\(\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}.\left(11-3\right)}{2^2.3^{28}}=\frac{3^{28}.3.2^3}{2^2.3^{28}}=6\)
\(=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}=\frac{11.3^{29}-3^{30}}{4.3^{28}}=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}=\frac{3.8}{4}=6\)