How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$?

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

We have

$(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz$

With $x=9\sqrt[3]{9},y=-3\sqrt[3]{3},z=-27$ all terms on the right side are rational, try it. So multiply the given numerator and denominator by $x^2+y^2+z^2-xy-xz-yz$ with $x,y,z$ as rendered above.

Compute the minimal polynomial of $\alpha=-27-3\cdot3^{1/3}+9\cdot3^{2/3}$: $$ \begin{align} \alpha&=-27-3\cdot3^{1/3}+9\cdot3^{2/3}\tag1\\ \alpha^2&=567+405\cdot3^{1/3}-477\cdot3^{2/3}\tag2\\ \alpha^3&=-81-25515\cdot3^{1/3}+16767\cdot3^{2/3}\tag3 \end{align} $$ Since $(-3,405,-25515)\times(9,-477,16767)=-2214(1,81,2430)$, we get that $$ \alpha^3+81\alpha^2+2430\alpha+19764=0\tag4 $$ Thus, dividing $(4)$ by $19764\alpha$, we get $$ \begin{align} \frac1\alpha &=-\frac{\alpha^2+81\alpha+2430}{19764}\\ &=-\frac{45+9\cdot3^{1/3}+14\cdot3^{2/3}}{1098}\tag5 \end{align} $$