Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
add comment

3 Answers 3

up vote 3 down vote accepted

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

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

share|improve this answer
add comment

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 $$\frac{1}{2^{1/3}-2} = \frac{2^{2/3} + 2(2^{1/3}) + 4}{(2^{1/3}-2)(2^{2/3} + 2(2^{1/3})+4)} = \frac{2^{2/3} + 2^{4/3} + 4}{2 - 8} = -\frac{2^{2/3} + 2^{4/3} + 4}{6}. $$

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.