How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?

Let $x=3^\frac13$ then the fraction becomes
$$-\frac{1}{x^4}\cdot\frac{1}{x^5-x^4+1}$$
What we need is to multiply nominator and denominator with some polynomial of $x$ in order to leave only terms with exponents divisible by $3$ in the denominator. How can we find such polynomial?
The answer is simple: find remained when dividing a polynomial by $x^5-x^4+1$ and check if we can obtain such polynomial. Maybe it is confusing, here is an example.
Let's start with $x^6+ax^3+b$. Suppose we want to multiply $x^5-x^4+1$ with something to obtain $x^6+ax^3+1$ (because it will cancel out all cube roots). Here $a,b$ are constants. So, the remainder of the division is
$$x^4+ax^3-x+b-1$$
Then, check if we can chose $a,b$ in such a way to make this remainder $0$. If we can, then we are done. Otherwise check the next polynomial $x^9+ax^6+bx^3+c$. The ramainder is now
$$a x^4+(-a-1) x-a+(b-1) x^3+c-x^2-1$$
Can we chose some $a,b,c$ to make this $0$? No, because $a$ must be $0$ to cancel out $x^4$, but at the same time must be $-1$ to cancel out $x$. So, continue with checking polynomials. x^12+ax^9+bx^6+cx^3+d is the next one we want to check. Here is the remainder
$$x (-a-b+1)+x^3 (-a+c-1)-a x^2-a+(b-3) x^4-b+d+2$$
What we have is basically a system of equations
$$a+b=1\\a-c=-1\\-a=0\\b-3=0\\a-b-d=2$$
This system of equations has no solutions. Let's continue. Remainder for degree $15$
$$x (a-b-c+4)+x^3 (-a-b+d+2)+x^4 (-3 a+c-3)+2 a+(3-b) x^2-b-c+e+4$$
Good news! This system has a solution! We have
$$a=-1\\b=3\\c=0\\d=0\\e=1$$
which corresponds to the polynomial
$$x^{15} - x^{12} + 3 x^9 + 1$$
and we want to multiply denominator with
$$1 + x^4 - x^5 + x^8 + x^9 + x^{10}$$
Solution
As explained above, our polynomial is
$$1 + x^4 - x^5 + x^8 + x^9 + x^{10}$$
But, notice the $\frac1{x^4}$, so we need to multiply our polynomial with $x^2$.
Verification
Mathematica can verify that this actually works

