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 :)
 A: If the question is$\displaystyle\frac{1}{\sqrt[3]{2}-2}$ use formula $A^3-B^3=(A-B)(A^2+AB+B^2)$
$$\frac{1}{\sqrt[3]{2}-2}\frac{\sqrt[3]{2^2}+2\sqrt[3]{2}+4}{\sqrt[3]{2^2}+2\sqrt[3]{2}+4} =\frac{\sqrt[3]{4}+2\sqrt[3]{2}+4}{-6}$$
A: 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
$$\frac{1}{2^{1/3}-2} = \frac{2^{2/3} + 2(2^{1/3}) + 4}{(2^{1/3}-2)(2^{2/3} + 2(2^{1/3})+4)} = \frac{2^{2/3} + 2^{4/3} + 4}{2 - 8} = -\frac{2^{2/3} + 2^{4/3} + 4}{6}. $$
A: 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.
