Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$

I keep making a mess of this. I tried vewing the denominator as

$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as

$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.

Then using the sum and differences in cubes fratorization but this keeps adding radicals to the denominator.

How should I approach this/where could I be going wrong?

## Best Answer

$9-3\sqrt[3]{3}+\sqrt[3]{9} = a^2 -ab+b^2$

$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}} \cdot \frac{3+\sqrt[3]{9}}{3+\sqrt[3]{9}}=\frac{12+4\sqrt[3]{9}}{30}$