Answer
$\frac{9}{4x^{\frac{1}{3}}}$
Work Step by Step
First, we will simplify the expression within the parenthesis using the rule $\frac{a^{m}}{a^{n}}=a^{m-n}$:
$(\frac{3x^{\frac{1}{3}}}{2x^{\frac{1}{2}}})^{2}$
=$(\frac{3}{2}\times\frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}})^{2}$
=$(\frac{3}{2}\times x^{\frac{1}{3}-\frac{1}{2}})^{2}$
=$(\frac{3}{2}\times x^{\frac{2-3}{6}})^{2}$
=$(\frac{3}{2}\times x^{\frac{-1}{6}})^{2}$
Now, we raise each term in the expression to the power of $2$:
$(\frac{3}{2}\times x^{\frac{-1}{6}})^{2}$
=$(\frac{3}{2})^{2}\times (x^{\frac{-1}{6}})^{2}$
=$(\frac{3^{2}}{2^{2}})\times (x^{\frac{-2}{6}})$
=$(\frac{9}{4})\times (x^{\frac{-1}{3}})$
=$(\frac{9}{4})\times \frac{1}{x^{\frac{1}{3}}}$
=$\frac{9}{4x^{\frac{1}{3}}}$
