# Math Help - Rdical 1

1. ## Rdical 1

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

2. Originally Posted by dhiab
Prove :
$
\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}
$
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

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

3. looks like Ramanujan's identity . . . .