Rdical 1

**dhiab** · November 12th 2009, 10:34 AM

Prove :
$<br /> \sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}<br />$

**Opalg** · November 12th 2009, 12:09 PM

Originally Posted by dhiab

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

$\Bigl(\frac{1-\omega+\omega^2}{\sqrt[3]9}\Bigr)^3 = \frac{(1+2)^2} {9(\omega+1)^3} = \frac3{(\omega+1)^3} = \frac3{3(\omega^2+\omega+1)} = \frac{\omega-1}{\omega^3-1} = \omega-1.$

**pacman** · November 14th 2009, 04:17 AM

looks like Ramanujan's identity . . . .

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