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?
 A: $9-3\sqrt[3]{3}+\sqrt[3]{9} = a^2 -ab+b^2$
$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}} \cdot \frac{3+\sqrt[3]{9}}{3+\sqrt[3]{9}}=\frac{12+4\sqrt[3]{9}}{30}$
A: 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 $d=3$
A: To solve the general case, use the formula for the rationalization of 3 cube roots. $$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}=\frac{\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}-\sqrt[3]{ab}-\sqrt[3]{ac}-\sqrt[3]{bc}\right)\left(\left(3\sqrt[3]{abc}+a+b+c\right)^{2}-3\left(a+b+c\right)\sqrt[3]{abc}\right)}{\left(a+b+c\right)^{3}-27abc}$$ Your particular problem appears to be rigged such that the answer simplifies greatly. $$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}=\frac{2\sqrt[3]{3}+6}{15}$$
