which of the following is equivalent to $\frac{sqrt3{32x^{3}y^{6}}}{sqrt3{2x^{9}y^{2}}}$, where $xgeq0$ and…

which of the following is equivalent to $\frac{sqrt3{32x^{3}y^{6}}}{sqrt3{2x^{9}y^{2}}}$, where $xgeq0$ and $ygeq0$?\n$sqrt3{16x^{6}y^{4}}$\n$sqrt3{\frac{y^{4}}{16x^{6}}}$\n$sqrt3{\frac{16y^{4}}{x^{6}}}$

which of the following is equivalent to $\frac{sqrt3{32x^{3}y^{6}}}{sqrt3{2x^{9}y^{2}}}$, where $xgeq0$ and $ygeq0$?\n$sqrt3{16x^{6}y^{4}}$\n$sqrt3{\frac{y^{4}}{16x^{6}}}$\n$sqrt3{\frac{16y^{4}}{x^{6}}}$

Answer

Answer:

$\sqrt[3]{\frac{16y^{4}}{x^{6}}}$

Explanation:

Step1: Use quotient - rule of radicals

$\frac{\sqrt[3]{32x^{3}y^{6}}}{\sqrt[3]{2x^{9}y^{2}}}=\sqrt[3]{\frac{32x^{3}y^{6}}{2x^{9}y^{2}}}$

Step2: Simplify the coefficient and variables

Simplify the coefficient $\frac{32}{2}=16$. For the $x$ - terms, use the rule $\frac{x^{m}}{x^{n}}=x^{m - n}$, so $\frac{x^{3}}{x^{9}}=x^{3-9}=x^{- 6}=\frac{1}{x^{6}}$. For the $y$ - terms, $\frac{y^{6}}{y^{2}}=y^{6 - 2}=y^{4}$. Then $\sqrt[3]{\frac{32x^{3}y^{6}}{2x^{9}y^{2}}}=\sqrt[3]{\frac{16y^{4}}{x^{6}}}$