Rationalize the denominator of $\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$ Rationalize the denominator of $$\dfrac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$$ Usually we are supposed to use one of the formulas $$x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$$ I don't think they will work here. We can say $\sqrt[3]{3}=t\Rightarrow t^3=3$ and the given expression is then $$\dfrac{1}{1+t-t^2}$$ I don't see anything else. What are the available approaches?

You can proceed by identification
$(1+t-t^2)(a+bt+ct^2)=(b+c-a)t^2+(a+b-3c)t+(3c-3b+a)$
To get rid of the surds, solve $\begin{cases}b+c-a=0\\a+b-3c=0\end{cases}\implies\begin{cases}a=2c\\b=c\end{cases}$
We can set $c=1$ which correspond to the $(2+t+t^2)$ indicated in the other answer.
