divide and simplify: $\frac{sqrt3{48x^{3}y^{4}}}{sqrt3{3x^{2}}}$\n$ysqrt3{45xy}$\n$sqrt3{16xy^{4}}$\n$2ysqrt3…

Explanation:

Step1: Use quotient - rule of radicals

The quotient - rule states that $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$ for $a\geq0$, $b > 0$ and $n\geq2$. So, $\frac{\sqrt[3]{48x^{3}y^{4}}}{\sqrt[3]{3x^{2}}}=\sqrt[3]{\frac{48x^{3}y^{4}}{3x^{2}}}$.

Step2: Simplify the fraction inside the cube - root

Divide the coefficients and use the rule of exponents $\frac{x^{m}}{x^{n}}=x^{m - n}$. $\frac{48}{3}=16$, and $\frac{x^{3}}{x^{2}}=x^{3-2}=x$. So, $\sqrt[3]{\frac{48x^{3}y^{4}}{3x^{2}}}=\sqrt[3]{16xy^{4}}$.

Step3: Rewrite the radicand for further simplification

Rewrite $16xy^{4}$ as $8y^{3}\cdot2xy$. Then $\sqrt[3]{16xy^{4}}=\sqrt[3]{8y^{3}\cdot2xy}$.

Step4: Use the product - rule of radicals

The product - rule states that $\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}$ for $a\geq0$, $b\geq0$ and $n\geq2$. So, $\sqrt[3]{8y^{3}\cdot2xy}=\sqrt[3]{8y^{3}}\cdot\sqrt[3]{2xy}$.

Step5: Simplify $\sqrt[3]{8y^{3}}$

Since $\sqrt[3]{8y^{3}} = 2y$ (because $2^{3}=8$ and $(y^{1})^{3}=y^{3}$), then $\sqrt[3]{8y^{3}}\cdot\sqrt[3]{2xy}=2y\sqrt[3]{2xy}$.

Answer:

$2y\sqrt[3]{2xy}$