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Math Help - Rdical 1

  1. #1
    Super Member dhiab's Avatar
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    Rdical 1

    Prove :
     <br />
\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}<br />
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by dhiab View Post
    Prove :
     <br />
\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}<br />
    Let \omega = \sqrt[3]2. Then \sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}} = \frac{1-\omega+\omega^2}{\sqrt[3]9}. Use the identities \omega^3 + 1 = (\omega +1)(\omega^2-\omega+1), \omega^3 - 1 = (\omega -1)(\omega^2+\omega+1) and (\omega+1)^3 = 2 + 3\omega^2+3\omega+1 = 3(\omega^2+\omega+1) to see that

    \Bigl(\frac{1-\omega+\omega^2}{\sqrt[3]9}\Bigr)^3 = \frac{(1+2)^2} {9(\omega+1)^3} = \frac3{(\omega+1)^3} = \frac3{3(\omega^2+\omega+1)} = \frac{\omega-1}{\omega^3-1} = \omega-1.
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  3. #3
    Senior Member pacman's Avatar
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    looks like Ramanujan's identity . . . .
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