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Math Help - Radicals

  1. #1
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    Unhappy Radicals

    Can anyone solve this and explain it to me please. thank you!~!

    Sqrt of 3 / 1 - Ceubroot of 2 + (Sqrt of 3)(Cuberoot of 2 )

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  2. #2
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    Dear Anemori,

    Is this what you mean, \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear Anemori,

    Is this what you mean, \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})



    Yes sir, that's what i meant.
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  4. #4
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    Quote Originally Posted by Sudharaka View Post
    Dear Anemori,

    Is this what you mean, \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})
    Quote Originally Posted by Anemori View Post
    Yes sir, that's what i meant.
    Okay, then, what do you want to do with it? That is an "expression", not a "problem". Are you trying to add the two? To do that you will have to get fractions with the same denominator by multiplying \sqrt{3}\sqrt[3]{2} by the unit fraction \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Okay, then, what do you want to do with it? That is an "expression", not a "problem". Are you trying to add the two? To do that you will have to get fractions with the same denominator by multiplying \sqrt{3}\sqrt[3]{2} by the unit fraction \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}.

    Simplified it. can you explain how do you do it? i understand you cannot multiply or divide both radicals with different index.
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  6. #6
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    Dear Anemori,

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

    \frac{\sqrt{3}+(1-\sqrt[3]{2})\sqrt{3}\sqrt[3]{2}}{1-\sqrt[3]{2}}

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

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

    Hope this solves the questions you have about radical expressions.
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  7. #7
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    Hello, Anemori!

    Simplify: . \frac{\sqrt{3}}{1 - \sqrt[3]{2}} + \sqrt{3}\cdot\sqrt[3]{2}

    Factor out \sqrt{3}\!:\;\;\sqrt{3}\,\bigg[\frac{1}{1-\sqrt[3]{2}} + \sqrt[3]{2}\bigg]


    Multiply the fraction by \frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{1 + \sqrt[3]{2} + \sqrt[3]{4}}

    . . \frac{1}{1-\sqrt[3]{2}}\cdot\frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{1+\sqrt[3]{2} + \sqrt[3]{4}} \;=\; \frac{1+\sqrt[3]{2} + \sqrt[3]{4}}{1-2} \;=\;-\left(1 + \sqrt[3]{2} + \sqrt[3]{4}\right) . =\;-1-\sqrt[3]{2} - \sqrt[3]{4}


    The problem becomes: . \sqrt{3}\bigg[-1 - \sqrt[3]{2} - \sqrt[3]{4} + \sqrt[3]{2}\bigg] \;=\;\sqrt{3}\bigg[-1 - \sqrt[3]{4}\bigg]


    Therefore: . -\sqrt{3}\left(1 + \sqrt[3]{4}\right)

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