Answer
$ \displaystyle \frac{12^{3}}{y^{8}}$
Work Step by Step
$\displaystyle \frac{12^{3/4}\cdot 12^{5/4}\cdot y^{-2}}{12^{-1}\cdot(y^{-3})^{-2}}$=$\qquad$ .... group like terms
=$\displaystyle \frac{12^{3/4}\cdot 12^{5/4}}{12^{-1}}\cdot\frac{y^{-2}}{(y^{-3})^{-2}} \qquad$
....... use $a^{m}\cdot a^{n}=a^{m+n}$ and $(a^{m})^{n}=a^{mn}$
=$\displaystyle \frac{12^{3/4+5/4}}{12^{-1}}\cdot\frac{y^{-2}}{y^{-3(-2)}}$
=$\displaystyle \frac{12^{2}}{12^{-1}}\cdot\frac{y^{-2}}{y^{6}}\qquad$ ....... use $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$
$=12^{2-(-1)}\cdot y^{-2-6}$
$=12^{3}y^{-8}$
$= \displaystyle \frac{12^{3}}{y^{8}}$
