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Math Help - Quotient? I'm having a tough time with it would appreciate help!

  1. #1
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    Quotient? I'm having a tough time with it would appreciate help!

    Simplify using the product and quotient properties of radicals

    cube root of 4X^4/25Y^6
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  2. #2
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    \displaystyle \sqrt[3]{\frac{4x^4}{25y^6}} = \frac{\sqrt[3]{4x^4}}{\sqrt[3]{25y^6}}

    \displaystyle = \frac{\sqrt[3]{4}\sqrt[3]{x^4}}{\sqrt[3]{25}\sqrt[3]{y^6}}

    \displaystyle = \frac{\sqrt[3]{4}\sqrt[3]{x^4}}{\sqrt[3]{25}\,y^2}

    \displaystyle = \frac{\sqrt[3]{5}\sqrt[3]{4}\sqrt[3]{x^4}}{\sqrt[3]{5}\sqrt[3]{5^2}\,y^2}

    \displaystyle = \frac{\sqrt[3]{5\cdot 4\cdot x^4}}{\sqrt[3]{5^3}\,y^2}

    \displaystyle = \frac{\sqrt[3]{20x^4}}{5y^2}.
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  3. #3
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    Quote Originally Posted by victorfk06 View Post
    Simplify using the product and quotient properties of radicals

    cube root of 4X^4/25Y^6
    You are asked to calculate partially the cube root of a term:

    \sqrt[3]{\dfrac{4x^4}{25y^6}} = \sqrt[3]{\dfrac{4x^4 \cdot 2 \cdot 5}{25y^6 \cdot 2 \cdot 5}} =

    Now re-arrange:

    \sqrt[3]{\dfrac{2^3 x^3 \cdot 5x}{5^3 y^6 \cdot 2 }} =

    Can you take it from here?
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    Quote Originally Posted by earboth View Post
    You are asked to calculate partially the cube root of a term:

    \sqrt[3]{\dfrac{4x^4}{25y^6}} = \sqrt[3]{\dfrac{4x^4 \cdot 2 \cdot 5}{25y^6 \cdot 2 \cdot 5}} =

    Now re-arrange:

    \sqrt[3]{\dfrac{2^3 x^3 \cdot 5x}{5^3 y^6 \cdot 2 }} =

    Can you take it from here?
    Of course, that will leave you with an irrational denominator...
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  5. #5
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    Hello, victorfk06!

    \text{Simplify using  the product and quotient properties of radicals:}

    . . \sqrt[3]{\dfrac{4x^4}{25y^6}}

    \displaystyle \text{We have: }\;\sqrt[3]{\frac{4\cdot x^3 \cdot x}{5^2\cdot y^6}}


    \displaystyle \text{Under the radical multiply by }\frac{5}{5}\!:\;\;\sqrt[3]{\frac{5}{5}\cdot \frac{4\cdot x^3 \cdot x}{5^2\cdot y^6}}


    \displaystyle \text{We have: }\; \sqrt[3]{\frac{20\cdot x^3 \cdot x}{5^3\cdot y^6}} \;=\;\frac{\sqrt[3]{20}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{x}} {\sqrt[3]{5^3}\cdot\sqrt[3]{y^6}}

    . . . . . . . . \displaystyle =\;\frac{\sqrt[3]{20}\cdot x \cdot \sqrt[3]{x}}{5\cdot y^2} \;=\;\frac{x\sqrt[3]{20x}}{5y^2}

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