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?

Notice that $(1 + t - t^2)(2 + t + t^2) = 2 + 3t - t^4$. Since $t^4 = 3t$, this implies
$$(1 + t - t^2)(2 + t + t^2) = 2.$$
Hence
$$\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}} = \frac{1}{2}(2 + \sqrt[3]{3} + \sqrt[3]{9}).$$
EDIT: As to Mark's comment: You know it must be something of the form $a + bt + t^2$. To get rid of the linear and quadratic term, we must have $a+b=3$ and $-a+b=-1$, which easily gives $a = 2$ and $b = 1$.
