what is $\frac{(x^{2}y^{3})^{\frac{1}{3}}}{sqrt3{x^{2}y}}$ in exponential form?\n$\frac{x^{\frac{2}{3}}y}{x^{…

Explanation:

Step1: Rewrite radicals as exponents

Recall that $\sqrt[n]{a}=a^{\frac{1}{n}}$. So, $(x^{2}y^{3})^{\frac{1}{3}}=x^{\frac{2}{3}}y$ and $\sqrt[3]{x^{2}y}=x^{\frac{2}{3}}y^{\frac{1}{3}}$.

Step2: Use the rule of dividing exponents

When dividing terms with the same base $a^m\div a^n=a^{m - n}$. So, $\frac{x^{\frac{2}{3}}y}{x^{\frac{2}{3}}y^{\frac{1}{3}}}=x^{\frac{2}{3}-\frac{2}{3}}y^{1-\frac{1}{3}}=x^{0}y^{\frac{2}{3}} = y^{\frac{2}{3}}$. But let's start over and do it in a more general - way for the whole expression. First, $(x^{2}y^{3})^{\frac{1}{3}}=x^{\frac{2}{3}}y$ and $\sqrt[3]{x^{2}y}=x^{\frac{2}{3}}y^{\frac{1}{3}}$. Then $\frac{(x^{2}y^{3})^{\frac{1}{3}}}{\sqrt[3]{x^{2}y}}=\frac{x^{\frac{2}{3}}y}{x^{\frac{2}{3}}y^{\frac{1}{3}}}=x^{\frac{2}{3}-\frac{2}{3}}y^{1 - \frac{1}{3}}=y^{\frac{2}{3}}$. If we consider the general form of exponent - rules: [ \begin{align*} \frac{(x^{2}y^{3})^{\frac{1}{3}}}{\sqrt[3]{x^{2}y}}&=\frac{x^{\frac{2}{3}}y}{x^{\frac{2}{3}}y^{\frac{1}{3}}}\ &=x^{\frac{2}{3}-\frac{2}{3}}y^{1-\frac{1}{3}}\ &=x^{0}y^{\frac{2}{3}} \end{align*} ] The correct option is the first one $\frac{x^{\frac{2}{3}}y}{x^{\frac{2}{3}}y^{\frac{1}{3}}}$.

Answer:

The first option: $\frac{x^{\frac{2}{3}}y}{x^{\frac{2}{3}}y^{\frac{1}{3}}}$