1
$\begingroup$

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 :)

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ the second one is what i meant, thanks :) $\endgroup$ – Bartlomiej Lewandowski Oct 2 '11 at 20:45
  • $\begingroup$ 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. $\endgroup$ – Bartlomiej Lewandowski Oct 2 '11 at 20:50
  • $\begingroup$ and a third one, the formula is equal to a^3 - b^3 $\endgroup$ – Bartlomiej Lewandowski Oct 2 '11 at 20:51
  • $\begingroup$ 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 $(2^{1/3})^2+(2^{1/3})2 + 2^2 = 2^{2/3}+2(2^{1/3})+4$. $\endgroup$ – Arturo Magidin Oct 2 '11 at 20:54
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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