Timeline for Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12 at 13:25 | history | edited | Michael Rozenberg | CC BY-SA 4.0 | added 1228 characters in body |
| Aug 3, 2019 at 5:03 | comment | added | Michael Rozenberg | @Frank W Such that $$A^3B^3=uA^2B^2+vAB+w$$ would be solvable equation like this: $A^3B^3=9A^2B^2-27AB+20$ | |
| Aug 3, 2019 at 4:56 | comment | added | Frank W | @MichaelRozenberg How do you determine the $p, q, r$ then? | |
| Jul 29, 2019 at 4:42 | comment | added | Michael Rozenberg | @Frank W It does not work in the general cases, otherwise identities as Ramanudjan's stile not would be so interesting. | |
| Jul 29, 2019 at 4:20 | comment | added | Frank W | @MichaelRozenberg Can you explain a bit more on how you proved your identity? I followed your comment and arrived at $$\begin{align*}A^3 & =3AB+3\sqrt[3]{r}-p\\B^3 & =-3\sqrt[3]{r}AB-3\sqrt[3]{r^2}+q\end{align*}$$So that$$z^3=-(3z+3\sqrt[3]{r}-p)(3z\sqrt[3]{r}+3\sqrt[3]{r^2}-q)$$But how do I solve for $z=AB$? Whenever I plug into WA, it spits out a messy solution... | |
| Feb 28, 2019 at 3:50 | comment | added | stressed out | Maybe it's a little bit irrelevant, but how did you remember this? Is there an online repository or search engine for this sort of stuff or you recited it from your memory?! | |
| Apr 20, 2017 at 5:27 | comment | added | Pragyaditya Das | @MichaelRozenberg Thanks a lot for this. Is there any online source I can look upto? | |
| Apr 19, 2017 at 10:17 | comment | added | Michael Rozenberg | @Pragyaditya Das Yes, of course! Let $a$, $b$ and $c$ be real roots of the equation $x^3+px^2+qx+r=0$. Also Let $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c=A$ and $\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}=B$. Hence, try to find $p$, $q$ and $r$ such that $A^3=kAB+n$ and $B^3=mAB+l$ solve the equation $x^3=(kx+n)(mx+l)$. | |
| Apr 19, 2017 at 8:28 | comment | added | Pragyaditya Das | @MichaelRozenberg Can you direct me to the proof? | |
| Apr 19, 2017 at 8:26 | comment | added | Michael Rozenberg | @Pragyaditya Das Yes, there is an algebraic proof. | |
| Apr 19, 2017 at 8:25 | comment | added | Pragyaditya Das | @MichaelRozenberg Is there any algebraic proof for this? | |
| Apr 19, 2017 at 8:24 | comment | added | Pragyaditya Das | Can I have a proof for this identity? A possible link or a book? | |
| May 9, 2016 at 3:19 | comment | added | Michael Rozenberg | @Frank your calculation. | |
| Mar 26, 2016 at 4:57 | comment | added | Tito Piezas III | +1 Very nice. Note that the LHS can be simplified as $$\sqrt[3]{(m-n)^3+9m^2n-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}$$ | |
| Mar 26, 2016 at 4:57 | history | edited | Tito Piezas III | CC BY-SA 3.0 | Formatting. |
| Mar 26, 2016 at 4:21 | history | answered | Michael Rozenberg | CC BY-SA 3.0 |