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?

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.
