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?

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. $$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}=\frac{2\sqrt[3]{3}+6}{15}$$
