Answer
$\dfrac{8}{x^{3}y}$
Work Step by Step
Use $4=2^2$.
$\dfrac{(4x^{-1}y^{\frac{1}{3}})^{\frac{3}{2}}}{(xy)^{\frac{3}{2}}}=\dfrac{(2^2x^{-1}y^{\frac{1}{3}})^{\frac{3}{2}}}{(xy)^{\frac{3}{2}}}$
Use the rule $\left(ab\right)^m=a^mb^m$:
$=\dfrac{\left(2^2\right)^{\frac{3}{2}}\left(x^{-1}\right)^{\frac{3}{2}}\left(y^{\frac{1}{3}}\right)^{\frac{3}{2}}}{x^{\frac{3}{2}}y^{\frac{3}{2}}}$
Use the rule $\left(a^m\right)^n=a^{mn}$, then simplify to obtain:
$=\dfrac{2^{2\cdot \frac{3}{2}}x^{-1\cdot \frac{3}{2}}y^{\frac{1}{3}\cdot \frac{3}{2}}}{x^{\frac{3}{2}}y^{\frac{3}{2}}}$
$=\dfrac{2^3x^{ -\frac{3}{2}}y^{\frac{1}{2}}}{x^{\frac{3}{2}}y^{\frac{3}{2}}}$
$=\dfrac{8x^{ -\frac{3}{2}}y^{\frac{1}{2}}}{x^{\frac{3}{2}}y^{\frac{3}{2}}}$
Use the rule $\dfrac{a^m}{a^n} = a^{m-n}$, then simplify to obtain:
$=8x^{-\frac{3}{2}-\frac{3}{2}}y^{\frac{1}{2}-\frac{3}{2}}$
$=8x^{-3}y^{-1}$
Use the rule $a^{-m}=\dfrac{1}{a^m}$, then simplify to obtain:
$=8 \cdot \dfrac{1}{x^3} \cdot \dfrac{1}{y}$
$=\dfrac{8}{x^3y}$
Hence, the correct answer is $\frac{8}{x^{3}y}$.
