# Getting the root out of the denominator

Alright quick question,

If I have

$$\frac{1}{2^{1/3} - 2},$$

how do i change it to not have the root in the denominator? I know the trick with the root of 2, but this seems a bit more complicated. Thanks in advance :)

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If the question is$\displaystyle\frac{1}{\sqrt[3]{2}-2}$ use formula $A^3-B^3=(A-B)(A^2+AB+B^2)$

$$\frac{1}{\sqrt[3]{2}-2}\frac{\sqrt[3]{2^2}+2\sqrt[3]{2}+4}{\sqrt[3]{2^2}+2\sqrt[3]{2}+4} =\frac{\sqrt[3]{4}+2\sqrt[3]{2}+4}{-6}$$

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If you mean $$\frac{1}{2^{1/3}} - 2,$$ then just multiply the numerator and denominator by $2^{2/3}$; we get $$\frac{2^{2/3}}{2} - 2.$$

If you mean $$\frac{1}{2^{1/3} - 2}$$ then you can use the formula $$(a-b)(a^2+ab+b^2) = a^3-b^3$$ so $$\frac{1}{2^{1/3}-2} = \frac{2^{2/3} + 2(2^{1/3}) + 4}{(2^{1/3}-2)(2^{2/3} + 2(2^{1/3})+4)} = \frac{2^{2/3} + 2^{4/3} + 4}{2 - 8} = -\frac{2^{2/3} + 2^{4/3} + 4}{6}.$$

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the second one is what i meant, thanks :) – Bartlomiej Lewandowski Oct 2 '11 at 20:45
i think there are two mistakes there, there isn't a 2 in the a^2+2ab+b^2 and you made a + instead of a minus in there as well. – Bartlomiej Lewandowski Oct 2 '11 at 20:50
and a third one, the formula is equal to a^3 - b^3 – Bartlomiej Lewandowski Oct 2 '11 at 20:51
Yes, it equals $a^3-b^3$; what "+ instead of a -"? $(a-b)(a^2+ab+b^2) = a^3+a^2b+ab^2 - a^2b-ab^2 - b^3 = a^3-b^3$, which is what I want. I'm applying the formula with $a=2^{1/3}$ and $b=2$, so $a^2+ab+b^2$ is equal to $(2^{1/3})^2+(2^{1/3})2 + 2^2 = 2^{2/3}+2(2^{1/3})+4$. – Arturo Magidin Oct 2 '11 at 20:54

HINT $\rm\ \ f(\alpha) = 0\ \Rightarrow\ (\alpha-n)\ \dfrac{f(\alpha)-f(n)}{\alpha-n}\ =\ -f(n)$

Thus $\rm\ \alpha^3 - 2\: =\: 0\ \Rightarrow (\alpha-2)\ \dfrac{\alpha^3-8}{\alpha-2}\ =\ {-}6\:,\:\$ i.e. $\rm\:\ (\alpha-2)\ (\alpha^2+2\ \alpha + 4)\ =\ {-}6$

This method works generally to invert any algebraic irrational $\rm\:\alpha - n\:$ given any integer coefficient polynomial $\rm\:f(x)\:$ such that $\rm\:f(\alpha) = 0\:\ne\: f(n)\:,\:$ e.g. choose $\rm\:f(x)\: =\:$ the minimal polynomial of $\rm\:\alpha\:.\:$ For further discussion see my posts on rationalizing denominators.

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