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

A

Explanation:

Answer 2

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

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* 3}{\sqrt[3]{3^2}}-3(\frac{\sqrt[3]{3* 2}}{\sqrt[3]{3^2}})`

`2*\frac{3}{3^{(2)/(3)}}-3(\frac{2^{(1)/(3)}* 3^{(1)/(3)}}{3^{(2)/(3)}})`

Using the property

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

`2* 3^{1-(2)/(3)}-3(2^{(1)/(3)}* 3^{(1)/(3)-(2)/(3)})`

Using the property

`(a^x)/(a^y)=a^(x-y)`

`2(3^{(1)/(3)})-3(2^{(1)/(3)}* 3^{-(1)/(3)})`

`2(3^{(1)/(3)})-2^{(1)/(3)}* 3^{1-(1)/(3)}`

`2(3^{(1)/(3)})-2^{(1)/(3)}* 3^{(2)/(3)}`

`2(3^{(1)/(3)})-2^{(1)/(3)}* \sqrt[3]{3^2}`

`2(3^{(1)/(3)})-2^{(1)/(3)}* \sqrt[3]{9}`

`2(3^{(1)/(3)})-\sqrt[3]{2* 9}`

`2(3^{(1)/(3)})-\sqrt[3]{18}`

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

`2(3^{(1)/(3)})-\sqrt[3]{18}`

Option A is correct.