what is the following quotient?\n$\frac{6 - 3(sqrt3{6})}{sqrt3{9}}$\n$2(sqrt3{3})-sqrt3{18}$\n$2(sqrt3{3})-3(…

Explanation:

Step1: Split the fraction

We use the rule $\frac{a - b}{c}=\frac{a}{c}-\frac{b}{c}$. So, $\frac{6 - 3\sqrt[3]{6}}{\sqrt[3]{9}}=\frac{6}{\sqrt[3]{9}}-\frac{3\sqrt[3]{6}}{\sqrt[3]{9}}$.

Step2: Rationalize the denominators

For $\frac{6}{\sqrt[3]{9}}$, we multiply the numerator and denominator by $\sqrt[3]{3}$ since $9\times3 = 27=3^3$. We get $\frac{6\sqrt[3]{3}}{\sqrt[3]{27}}=\frac{6\sqrt[3]{3}}{3}=2\sqrt[3]{3}$. For $\frac{3\sqrt[3]{6}}{\sqrt[3]{9}}$, we use the rule $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}=\sqrt[3]{\frac{a}{b}}$. So $\frac{3\sqrt[3]{6}}{\sqrt[3]{9}}=3\sqrt[3]{\frac{6}{9}}=3\sqrt[3]{\frac{2}{3}}$. Multiply the numerator and denominator inside the cube - root by $3^2$ to rationalize: $3\sqrt[3]{\frac{2\times3^2}{3^3}}=3\sqrt[3]{\frac{18}{27}} = 3\sqrt[3]{2}\times\frac{1}{ \sqrt[3]{27}}=3\sqrt[3]{2}\times\frac{1}{3}=\sqrt[3]{18}$.

Step3: Combine the results

$\frac{6}{\sqrt[3]{9}}-\frac{3\sqrt[3]{6}}{\sqrt[3]{9}}=2\sqrt[3]{3}-\sqrt[3]{18}$.

Answer:

$2(\sqrt[3]{3})-\sqrt[3]{18}$