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Thread: prove that

  1. September 6th 2008, 03:17 PM #1
    perash
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    prove that

    prove that

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  2. September 6th 2008, 08:23 PM #2
    Soroban
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    Hello, perash!

    I got the first half . . .

    \text{Let: }\:A \;=\;\sqrt[3]{3 + 12\sqrt[3]{11} - 6\sqrt[3]{121}}\;\text{ and }\;B \;=\;\sqrt[3]{16 - 27\sqrt[3]{11} + 9\sqrt[3]{121}}


    \text{Show (without using a calculator) that:}

    . . (a)\;\;A + B \:=\:1\qquad\qquad (b)\;\;A\cdot B \:< \:\frac{1}{4}

    I happened to notice that: . \begin{array}{ccc}3 + 12\sqrt[3]{11} - 6\sqrt[3]{121} &=& \left(\sqrt[3]{11} - 2\right)^3 \\ \\[-3mm] 16 -27\sqrt[3]{11} + 9\sqrt[3]{121} &=& \left(3 - \sqrt[3]{11}\right)^3 \end{array}


    (a) Therefore: . A + B \;=\;\sqrt[3]{\left(\sqrt[3]{11}-2\right)^3} + \sqrt[3]{\left(3-\sqrt[3]{11}\right)^3} \;=\;\left(\sqrt[3]{11} - 2\right) + \left(3 - \sqrt[3]{11}\right) \;=\;1
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  3. September 7th 2008, 05:06 PM #3
    sportsfan93b
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    Quote Originally Posted by Soroban View Post
    Hello, perash!

    I got the first half . . .


    I happened to notice that: . \begin{array}{ccc}3 + 12\sqrt[3]{11} - 6\sqrt[3]{121} &=& \left(\sqrt[3]{11} - 2\right)^3 \\ \\[-3mm] 16 -27\sqrt[3]{11} + 9\sqrt[3]{121} &=& \left(3 - \sqrt[3]{11}\right)^3 \end{array}


    (a) Therefore: . A + B \;=\;\sqrt[3]{\left(\sqrt[3]{11}-2\right)^3} + \sqrt[3]{\left(3-\sqrt[3]{11}\right)^3} \;=\;\left(\sqrt[3]{11} - 2\right) + \left(3 - \sqrt[3]{11}\right) \;=\;1
    In your final solution wouldnt the cube route and the 3rd pwr cancel out?
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  4. September 7th 2008, 05:14 PM #4
    Jhevon
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    Quote Originally Posted by sportsfan93b View Post
    In your final solution wouldnt the cube root and the 3rd pwr cancel out?
    what do you mean? they did cancel out...
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  5. September 7th 2008, 05:14 PM #5
    11rdc11
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    What do you mean? It did cancel out in part A or did I misunderstand your question?

    Lol I was too slow.
    Last edited by 11rdc11; September 7th 2008 at 05:16 PM. Reason: I'm to slow
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« (Square root of 3) / 2 = 19.5 / F | Exponents »

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