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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?

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2 Answers 2

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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 $b = 1$.

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    $\begingroup$ 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. $\endgroup$
    – Mark Viola
    Sep 13, 2022 at 15:12
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    $\begingroup$ 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. $\endgroup$ Sep 13, 2022 at 15:17
  • $\begingroup$ @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. $\endgroup$
    – Klaus
    Sep 13, 2022 at 15:23
  • $\begingroup$ (+1) for the edits $\endgroup$
    – Mark Viola
    Sep 13, 2022 at 16:00
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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.

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