1. Good handwriting. 2. Great energy! 3. Great sound quality, and your voice is very effective for audio/video. 4. Well organized. It is deceptively hard to do math this complex on a video but you handled it like a champ! Good job, sir
I did it with basics only: ∛(∛2-1) = 1/∛(∛4+∛2+1) - cubic root conjugate. We want to get rid of the outer root, so our only hope is if the trinomial is a "perfect cube" of something - which it is: (∛2+1)³ = 3∛4+3∛2+3, so the very first shot is lucky and we get: 1/∛(∛4+∛2+1) = ∛3/(∛2+1). This implies: (∛2+1)(∛a+∛b+∛c) = ∛3. Now let's set a = 3x, b = 3y, c = 3z. Then: (∛2+1)(∛x+∛y+∛z) = 1 which is "almost" the sum of cubes expansion. If we let x = 4, y = -2, z = 1 we get (∛2+1)(∛x+∛y+∛z) = (∛2)³+1³ = 3, so dividing both sides by 3 we find x = 4/27, y = -2/27 and z = 1/27 so finally a = 4/9, b = -2/9, c = 1/9
This RHS is out of proof, being ab initio a real source of confusion. It is only a final of the LHS.So,...work must be done only in LHS. Anyway...the proof is very interesting. Newton is a smart teacher who explains with the necessary patience. Congratulation prof. Prime Newton.
If you bring the negative sign to the inside of an odd root you get the same answer, however, if you bring it to an even root you get an imaginary answer
Well... Wait a moment please, sir... :D We didn't have ANY constraints EXCEPT that all three values should be element of rationals... Am i right or is this justified? I have a drastically easier approach to that problem, you know... Since we have three variables, but only one defining equation... you know... ...we are entitled to chose arbitrary values for two of the variables, as long as the remaining part(s) of the expression keep the ability to express the desired result. Since all three expression parts on the RHS are equivalent in their value range and all provide equally for a sufficient range, ... I use my freedom of choice to set a = 0, b = 0, c = 2^(1/3) - 1. And am done. No rule broken. Task giver satisfied - i hope :D
@@PrimeNewtons The six different solutions are just that a, b, and c are interchangeable. The answer should have been given that (a,b,c) belongs to the set {1/9, -2/9, 4/9}.
Remember, x is not a variable here. It is a specific constant. So the equation is only true here because of that substitution made. Your name is glorious ✨️
Sir ,I am a math lover.and running78yrs using bifocal spects. I hope u made this video for those who watch it in their mmobiles. As I was a teacher by profession I appreciate ur attitude as if u teaching in a classroom. But it's very difficult to follow ur board work as it's not even visible in mobiles ,I sa,y. Please try to write a little bold . Thank u.😂 Pl
Hello Prime Newtons! Haha great video! Lets keep working together 👍👍👍
Is there any numerical method available to calculate gamma function of non analytical numbers like (1/3), (1/5) ?
Man, im a young brazilian boy and your videos are helping me a lot to learn maths and english. Please, never stop making those videos :).
1. Good handwriting. 2. Great energy! 3. Great sound quality, and your voice is very effective for audio/video. 4. Well organized. It is deceptively hard to do math this complex on a video but you handled it like a champ! Good job, sir
I am 61 years old and I graduated in engineering 40 years ago. I find your videos very refreshing and I am enjoying the mental exercise. Keep going!
I did it with basics only: ∛(∛2-1) = 1/∛(∛4+∛2+1) - cubic root conjugate. We want to get rid of the outer root, so our only hope is if the trinomial is a "perfect cube" of something - which it is: (∛2+1)³ = 3∛4+3∛2+3, so the very first shot is lucky and we get: 1/∛(∛4+∛2+1) = ∛3/(∛2+1). This implies: (∛2+1)(∛a+∛b+∛c) = ∛3. Now let's set a = 3x, b = 3y, c = 3z. Then: (∛2+1)(∛x+∛y+∛z) = 1 which is "almost" the sum of cubes expansion. If we let x = 4, y = -2, z = 1 we get (∛2+1)(∛x+∛y+∛z) = (∛2)³+1³ = 3, so dividing both sides by 3 we find x = 4/27, y = -2/27 and z = 1/27 so finally a = 4/9, b = -2/9, c = 1/9
I am going to write this out and try it on other problems. Thanks for sharing 👍
Very intesting
Also came from PK Math. You guys should do this more often. You two both seem to be serious about math contents, which I love
Masterful. And fun at the same time.
This was a great problem! Enjoyed this a lot. Thanks Prof. Newtons!!
Came from PK Math. I was wowed, nice to see you guys collaborate on a problem. Btw, PKMath1234 on your title is not clickable.
No hesitation, added subscriber here 🙏🏻 teacher
Amazing Sir. Thank you forcñ this wonderful video
WOW! "never stop learning"... fantastic
I come from office tired and drained out. I watched this video and boom my mind was blown. Though in a great way. Thanks for making this video
Thank you, your work is very helpful
Excellent approach. A really nice problem
What a beatiful way to solve the problem!!
I am a math lover and your videos are amazing.
Nice dear friend.
I like to see you solving more STATISTICS problem (not probability question), eg distributions or testing type of questions.
very very respect
nice one
Your best video! Thank you
Fantástico! Aprendo muito aqui no seu canal. Excelente vídeo.
Great video! Thanks.
Just wonderful!!!! I love it.....
“Those who stop learning, stop living”
Great quote❤
Fantastic
Loved this one sir
This RHS is out of proof, being ab initio a real source of confusion. It is only a final of the LHS.So,...work must be done only in LHS. Anyway...the proof is very interesting. Newton is a smart teacher who explains with the necessary patience. Congratulation prof. Prime Newton.
it's magic! I simply love how algebra works
a nice jobe,thanks.
Amazing!!!!
Tremendous! 🎓
Good 👍
Vous etes formidable !!!!!
Thank for you
Awesome!!!
Very nice
Brilliant stuff! I'll watch it two or three times then I can pretend that I generated the answer myself.
😂
We all generate answers by ourselves once we have learned from others. The fastest way to learn.
Yes, a true society when we all genuinely learn from each other.
👏👏👏👏👍
I like your lesson!
I like math more
Just Nice!!
Beautiful
Here we go! Made my day bro)
nice
Can you solve it?! For what values of x and y, the equation is fulfilled: cos(x-y)=cosx-cosy ?
I liked this video ❤
I want a t-shirt. Do you sell them?
Prime Newtons is my hero! ❤🎉😊
Oh my... It kills my brain watching this because I still don't know how'd you get to those steps, literally all
very nice presentation and explanation to the problem. Does anybody knows from what year this USAMO problem is coming from ?
sir can we add negative mark in side the cubic root value when we inside it it becomes imaginary
If you bring the negative sign to the inside of an odd root you get the same answer, however, if you bring it to an even root you get an imaginary answer
Well... Wait a moment please, sir... :D
We didn't have ANY constraints EXCEPT that all three values should be element of rationals... Am i right or is this justified?
I have a drastically easier approach to that problem, you know...
Since we have three variables, but only one defining equation... you know...
...we are entitled to chose arbitrary values for two of the variables, as long as the remaining part(s) of the expression keep the ability to express the desired result.
Since all three expression parts on the RHS are equivalent in their value range and all provide equally for a sufficient range, ...
I use my freedom of choice to set a = 0, b = 0, c = 2^(1/3) - 1. And am done. No rule broken. Task giver satisfied - i hope :D
Sadly, this doesn't count because the problem asks for rational a, b and c. And cbrt(2)-1 is not rational man, but it was a nice idea :)
very good but there's 6 diferent solutions for (a,b,c)
I'd be happy to see your solution. Please send me an email. primenewtons@gmail.com
@@PrimeNewtons The six different solutions are just that a, b, and c are interchangeable. The answer should have been given that (a,b,c) belongs to the set {1/9, -2/9, 4/9}.
my question how we can transfer negative to under root cube????
Odd roots don't give complex numbers like even roots do.
@@henrybright6858 sorry i did not understand will you elaborate it ?
How is x^2+x+1 equal to 1/3(x+1)^3 pls Prof. Newton help me out.
Remember, x is not a variable here. It is a specific constant. So the equation is only true here because of that substitution made. Your name is glorious ✨️
@@PrimeNewtonsglorious 😂😂
Sir ,I am a math lover.and running78yrs using bifocal spects.
I hope u made this video for those who watch it in their mmobiles.
As I was a teacher by profession I appreciate ur attitude as if u teaching in a classroom.
But it's very difficult to follow ur board work as it's not even visible in mobiles ,I sa,y.
Please try to write a little bold . Thank u.😂
Pl
I watch on my mobile just fine.
Dude it's fine on mobile
Try putting it full screen horizontal mode.
Incredibly convoluted solution!
Magic
Great teacher
2, -1, 0
🇮🇹
これは気づかん
До степени перемешения
Hah
X cude? Or X cube... Or X cute?)))))
Very good
Beautiful