What is the following quotient? StartFraction 6 minus 3 (RootIndex 3 StartRoot 6 EndRoot) Over RootIndex 3 StartRoot 9 EndRoot E

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)

in progress 0
RuslanHeatt 4 years 2021-08-12T20:47:12+00:00 1 Answers 32 views 0

Answers ( )

  1. thuthuy
    0
    2021-08-12T20:48:12+00:00
    Reply

    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.

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