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

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?

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

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

in general we may factor the norm form $ x^3 + d y^3 + d^2 z^3 - 3dxyz$ as

$$ x^3 + d y^3 + d^2 z^3 - 3dxyz = $$ $$ \left( x+ d^{1/3}y + d^{2/3} z \right) \left( x^2 + d^{2/3} y^2 + d^{4/3} z^2 - dyz -d^{2/3}zx - d^{1/3} x y \right) $$

so that $$ \frac{1}{ x+ d^{1/3}y + d^{2/3} z} = $$ $$ \frac{x^2 + d^{2/3} y^2 + d^{4/3} z^2 - dyz -d^{2/3}zx - d^{1/3} x y}{ x^3 + d y^3 + d^2 z^3 - 3dxyz} $$

I guess it is desirable to write it according to the exponent of $d,$
$$ \frac{1}{ x+ d^{1/3}y + d^{2/3} z} = $$ $$ \frac{(x^2-dyz)+(dz^2 -xy) d^{1/3} + (y^2 - zx) d^{2/3} }{ x^3 + d y^3 + d^2 z^3 - 3dxyz} $$

For you $d=3$