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Math Help - Equation

  1. #1
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    Equation

    Hi people,

    I don't know how to calculate (\sqrt[3]{x}-1)^3-54=0???

    Can you help me please???
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  2. #2
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    Quote Originally Posted by bhitroofen01 View Post
    Hi people,

    I don't know how to calculate (\sqrt[3]{x}-1)^3-54=0???

    Can you help me please???
    Transfer all terms to the RHS except x:

    (\sqrt[3]{x}-1)^3-54=0

    (\sqrt[3]{x}-1)^3=54 Now calculate the 3rd root of both sides of the equation:

    \sqrt[3]{x}-1=\sqrt[3]{54}=3\sqrt[3]{2}

    \sqrt[3]{x}=1+3\sqrt[3]{2} Now get rid of the cube root:

    x=(1+3\sqrt[3]{2})^3
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  3. #3
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    Quote Originally Posted by bhitroofen01 View Post
    Hi people,

    I don't know how to calculate (\sqrt[3]{x}-1)^3-54=0???

    Can you help me please???
    (\sqrt[3]{x} - 1)^3 - 54 = 0

    (\sqrt[3]{x} - 1)^3 = 54

    \sqrt[3]{x} - 1 = \sqrt[3]{54}

    \sqrt[3]{x} - 1 = \sqrt[3]{27\cdot 2}

    \sqrt[3]{x} - 1 = \sqrt[3]{27} \cdot \sqrt[3]{2}

    \sqrt[3]{x} - 1 = 3\sqrt[3]{2}

    \sqrt[3]{x} = 3\sqrt[3]{2} + 1

    x = (3\sqrt[3]{2} + 1)^3.
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