i would luv it if you helped...

Simplify 

$\displaystyle \frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.

Geometric progression: $1 + \sqrt[3]3 + \sqrt[3]9 \qquad a = 1 \text{ and } r = \sqrt[3]3$

Formula: $1+r+r^2 = \dfrac{r^3-1}{r-1}$

$\begin{array}{|rcll|} \hline 1 + \sqrt[3]3 + \sqrt[3]9 &=& \dfrac{(\sqrt[3]3)^3-1}{\sqrt[3]3-1} \\\\ &=& \dfrac{3-1}{\sqrt[3]3-1} \\\\ &=& \dfrac{2}{\sqrt[3]3-1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9} } \\\\ &=& \dfrac{2\sqrt[3]9}{\dfrac{2}{\sqrt[3]3-1}} \\\\ &=& \sqrt[3]9(\sqrt[3]3-1) \\\\ &=& \sqrt[3]{3^3}-\sqrt[3]9 \\\\ &\mathbf{=}& \mathbf{3-\sqrt[3]9} \\ \hline \end{array}$