Question

What is the following quotient?

StartFraction 6 minus 3 (RootIndex 3 StartRoot 6 EndRoot) Over RootIndex 3 StartRoot 9 EndRoot EndFraction
2 (RootIndex 3 StartRoot 3 EndRoot) minus RootIndex 3 StartRoot 18 EndRoot
2 (RootIndex 3 StartRoot 3 EndRoot) minus 3 (RootIndex 3 StartRoot 2 EndRoot)
3 (RootIndex 3 StartRoot 3 EndRoot) minus RootIndex 3 StartRoot 18 EndRoot
3 (RootIndex 3 StartRoot 3 EndRoot) minus 3 (RootIndex 3 StartRoot 2 EndRoot)

Answer 1

Answer:

A.$2(3^{\frac{1}{3}})-\sqrt[3]{18}$

Step-by-step explanation:

We are given that

$\frac{6-3(\sqrt[3]{6}}{\sqrt[3]{9}})$

We have to find the quotient.

$\frac{6}{\sqrt[3]{9}}-3(\frac{\sqrt[3]{6}}{\sqrt[3]{9}})$

$\frac{2\times 3}{\sqrt[3]{3^2}}-3(\frac{\sqrt[3]{3\times 2}}{\sqrt[3]{3^2}})$

$2\times\frac{3}{3^{\frac{2}{3}}}-3(\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3^{\frac{2}{3}}})$

Using the property

$(ab)^n=a^n\cdot b^n$

$2\times 3^{1-\frac{2}{3}}-3(2^{\frac{1}{3}}\times 3^{\frac{1}{3}-\frac{2}{3}})$

Using the property

$\frac{a^x}{a^y}=a^{x-y}$

$2(3^{\frac{1}{3}})-3(2^{\frac{1}{3}}\times 3^{-\frac{1}{3}})$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times 3^{1-\frac{1}{3}}$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times 3^{\frac{2}{3}}$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times \sqrt[3]{3^2}$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times \sqrt[3]{9}$

$2(3^{\frac{1}{3}})-\sqrt[3]{2\times 9}$

$2(3^{\frac{1}{3}})-\sqrt[3]{18}$

Hence, the quotient of $\frac{6-3(\sqrt[3]{6}}{\sqrt[3]{9}})$ is given by

$2(3^{\frac{1}{3}})-\sqrt[3]{18}$

Option A is correct.

Answer 2

Answer:

A

Step-by-step explanation: