Rationalizing the denominator having square roots and cube roots In middle-school mathematics, the teachers always tell you that if you have radicals on the denominator of a fraction, then it isn't fit to be a final answer - you have to rationalize the denominator, or get rid of all of the radicals in the denominator by moving them to numerator.
Rationalizing the denominator is usually very easy, and can be done quickly using its conjugate. For example, consider
$$\frac{1}{2+\sqrt 2}$$
This denominator can be easily rationalized by using its conjugate:
$$\frac{2-\sqrt 2}{(2+\sqrt 2)(2-\sqrt 2)}$$
$$\frac{2-\sqrt 2}{4-2}$$
$$\frac{2-\sqrt 2}{2}$$
However, I have stumbled upon a new class of denominator-rationalization problems that I can't figure out how to solve. I was thoroughly stumped when I tried to rationalize the denominator of this fraction:
$$\frac{1}{2+\sqrt 2+\sqrt[3]{2}}$$
Can anybody figure out how to rationalize this? Is it even possible?
Or, more interestingly, if anyone suspects that it is not possible, how might one prove something like this?

Hint:  for a shortcut in this particular case, let $a = 2+\sqrt{2}$ then use that:
$$
\frac{1}{a+\sqrt[3]{2}} = \frac{a^2-a\sqrt[3]{2}+\sqrt[3]{4}}{a^3+2}
$$
The denominator now contains only integers and terms in $\sqrt{2}\,$ after expansion, which is the case you know how to rationalize.
