Question

Simplify:
$\frac{{\left(2^5\right)}^6\times {\left(3^3\right)}^2}{{\left(2^6\right)}^5\times {\left(3^2\right)}^3}$

Answer 1

$\frac{(2^5{)}^6\times {\left(3^3\right)}^2}{(2^6{)}^5\times {\left(3^2\right)}^3}$

$=\frac{(2^6{)}^5\times (3^2{)}^3}{(2^6{)}^5\times (3^2{)}^3}$

$=\frac{(2^6{)}^5\times (3^2{)}^3\times (3^2{)}^{-3}\times (2^6{)}^{-5}}{1}$

$=(2^6{)}^{5-5}\times (3^2{)}^{3-3}$

$=(2^6{)}^0\times (3^2{)}^0=1$

Hence, $\frac{(2^5{)}^6\times {\left(3^3\right)}^2}{(2^6{)}^5\times {\left(3^2\right)}^3}=1$