A Radical Approach To A Problem From ARML 😉

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[@SyberMath]

The original/book solution did not satisfy me. That's why I came up with my own version! 😊
ARML Power & Local Contests 2009-2014:
www.arml.com/ARML/arml_2019/public_contest_files/2009_2014_book/ARML_2009_2014.pdf

[@afrika-karibianaestudionan4050]

Greetings from Curaçao, an Island Nation in The Caribbean,
It seems like the provided link, does not have the "American Regions Mathematics League 1997" math question.

[@marksandsmith6778]

This gentleman's second language is English but he still knows the phrase " different from" and uses it correctly

[@SyberMath]

❤️

[@Qermaq]

Very impressive. A radical approach. ;)

[@SyberMath]

Your comment gave me an idea 😉

[@Qermaq]

@@SyberMath Uh-oh!

[@GlorifiedTruth]

Delightful! Another great video.

[@SyberMath]

Glad to hear that

[@scottleung9587]

Very good!

[@MrPoornakumar]

Brilliant ! I attempted a straight-forward approach of assuming "t" on RHS of eqn.. It doesn't work.

[@aliasgharheidaritabar9128]

Great problem

[@billtruttschel]

Skip to 2:50 for him to actually start working on the problem. You're welcome.

[@Jalina69]

Nice

[@yoav613]

Very nice

[@SyberMath]

Thanks

[@iSkyline1]

4:28 I don't get it, how do you know to multiply by the conjugate? I have a feeling this is the step that makes starting with the conjugate make sense...

[@leif1075]

Yea inthiught lf thr conjugate at the beginning but since we have a VUBE ROOT not SWUARE ROOT PF 2 I discarded it..did younthink lf that..and would you agree then using conjugate is kind of pitnof nowhere and most ppl if no one wluld.think of it??

[@NadiehFan]

This is just an application of the identity
(a − b)(a² + ab + b²) = a³ − b³
to get rid of the cube roots, exactly like you can use the identity
(a − b)(a + b) = a² − b²
to get rid of square roots. And how do we know to use this here? That is just a matter of pattern recognition. If you have
∛4 + ∛2 + 1
and you know that ∛4 = (∛2)² then you can see that we have
a² + ab + b²
with a = ∛2 and b = 1. Therefore, multiplying this by a − b = ∛2 − 1 will give a³ − b³ = (∛2)³ − 1³ = 2 − 1 = 1 so we have
(∛2 − 1)(∛4 + ∛2 + 1) = 1

[@anasharere]

One of the hardest questions I have ever seen

[@mystychief]

Why not = {12^(1/3) - 6^(1/3) + 3^(1/3)}/3 because in the forelast expression there is 1 denominator of the cube root of 27.

[@leif1075]

Syber thanks for a different approach but dont you ahree NO ONE WOULD.EVER THINK OF THIS METHOD so why lpresent this at all?? Woudlnt you ahree then its nkt eorth it snd depressing and naybe a cheat in thst case??

[@WhoAmIdotIn]

"Godzilla had a stroke trying to read that"

[@duongquocthongho2117]

you should have been more clear about the problem statement, "denesting" alone is quite vague

[@SyberMath]

I didn't want to give away the answer format

[@NadiehFan]

@@SyberMath I don't think there is anything 'vague' about _denesting_ as @duongquocthongho2117 claims. It is perfectly clear what is meant, express ∛(∛2 − 1) algebraically in a form that does not involve roots of expressions involving roots.
A century ago any highschool student knew what was meant by denesting nested roots, and denesting nested square roots was a standard part of the algebra curriculum. But this seems to have become something of a lost art in many quarters and some CZcams channels pass of the denesting of simple nested square roots as math Olympiad problems, which is quite ridiculous (and an indication of the sorry state of high school math education in much of the Western world).
But anyway it was nice to see a somewhat more challenging denesting problem from ARML. The result
∛(∛2 − 1) = ∛(¹⁄₉) − ∛(²⁄₉) + ∛(⁴⁄₉)
is attributed to Ramanujan who published it as a challenge in the Journal of the Indian Mathematical Society, but considering it is so elementary it is hard to believe it would not have been known earlier. He seems to have been fond of denesting problems because he came up with many more elaborate and peculiar results, for example
³√(⁵√(³²⁄₅) − ⁵√(²⁷⁄₅)) = ⁵√(¹⁄₂₅) + ⁵√(³⁄₂₅) − ⁵√(⁹⁄₂₅)
but I doubt we will see something like this in an ARML contest.