# 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 $$b = 1$$.

• You might consider adding the motivation for multiplying $1+t-t^2$ by $2+t+t^2$. That is, the choice does not seem obvious. Sep 13, 2022 at 15:12
• Another way is to make use of the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc).$ See this MSE answer, for example. Sep 13, 2022 at 15:17
• @MarkViola In this case I just played around with the coefficients, but you can always reduce it to a linear system. I edited the answer accordingly. Sep 13, 2022 at 15:23
• (+1) for the edits Sep 13, 2022 at 16:00

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.