Answer
$27 y^{\frac{2}{3}}$
Work Step by Step
RECALL:
(1) $a^m \cdot a^n = a^{m+n}$
(2) $\dfrac{a^m}{a^n} = a^{m-n}$
(3) $(ab)^m =a^mb^m$
Use rule (3) above to obtain:
$=\dfrac{3^3(y^{\frac{1}{4}})^3}{y^{\frac{1}{12}}}
\\=\dfrac{3(3)(3)(y^{\frac{1}{4}})^3}{y^{\frac{1}{12}}}
\\=\dfrac{27(y^{\frac{1}{4}})^3}{y^{\frac{1}{12}}}$
Use rule (3) above to obtain:
$=\dfrac{27(y^{\frac{1}{4}(3)})}{y^{\frac{1}{12}}}
\\=\dfrac{27y^{\frac{3}{4}}}{y^{\frac{1}{12}}}$
Use rule (2) above to obtain:
$=27 y^{\frac{3}{4} - \frac{1}{12}}$
Make the fractional exponents similar using their LCD of $12$ to obtain:
$=27 y^{\frac{9}{12} - \frac{1}{12}}
\\=27 y^{\frac{9-1}{12}}
\\=27 y^{\frac{8}{12}}
\\=27 y^{\frac{2}{3}}$
