what is the simplest form of $\frac{sqrt3{81x^{10}}}{sqrt3{3x}}$?\n3x\n3x³\n$3x^{4}sqrt{3x}$\n$3x^{3}sqrt3{x^…

Answer:

D. $3x^{3}\sqrt[3]{x^{2}}$

Explanation:

Step1: Use quotient - rule of radicals

$\frac{\sqrt[3]{81x^{10}}}{\sqrt[3]{3x}}=\sqrt[3]{\frac{81x^{10}}{3x}}$

Step2: Simplify the fraction inside the radical

$\sqrt[3]{\frac{81x^{10}}{3x}}=\sqrt[3]{27x^{9}}$

Step3: Rewrite and simplify

$\sqrt[3]{27x^{9}}=\sqrt[3]{27}\cdot\sqrt[3]{x^{9}}$. Since $\sqrt[3]{27} = 3$ and $\sqrt[3]{x^{9}}=x^{3}$, we have $3x^{3}\sqrt[3]{x^{2}}$ (because $x^{10}\div x=x^{9}$ and we can write $x^{9}\cdot x$ inside the original cube - root, and $\sqrt[3]{x^{9}\cdot x}=x^{3}\sqrt[3]{x}$).