Thank you for reminding us that there are, in general, six solutions to a 6th order equation. It would be more challenging to ask for all of the distinct solutions. Excellent video!.
Wow!!! what a wonderful prayer from a wonderful human/fellow from Gaza, Palestine. A loudest amen I and my team say to your awesome prayers sir. All of us @onlinemathstv love you and we also pray the Almighty God to protect and shine His face on you and your entire family members. Hope to meet with you one-on-one some days to come. We love you sir....💖💖💕💕❤️❤️😍😍
Excellent. But in Japan , Highschool students solve this question with complex plane. if n^3=2 , they think that n = r(cos𝜽 + isin𝜽) then other solution are 2^1/3(cos2/3𝛑+isin2/3𝛑) and 2^1/3(cos4/3𝛑+isin4/3𝛑), I think.All solution in this case exist 6 pattern.
It' s much simpler (just a couple of passages...) using Ruffini' s rule, once you recognize that n = -1 is one of the roots. The other real solution is cubic root of 2 (2^(1/3))
n^6-n^3=2 -2 from both sides let x=n^3 x^2-x-2=0 (one quadratic formula later) x=2 x=-1 x=n^3 => n=cbrt(-1)=-1 n=cbrt2 edit: ima try to find the imaginary roots just for fun x^6-x^3-2=0, if y=x^3, y^2-y-2=0 to factor y^2-y-2, 2 numbers (a & b) must satisfy a+b=-1 & ab=-2 =>a=-2 & b=1 => y^2-y-2=(y-2)(y+1) y=x^3, => x^6-x^3-2=(x^3-2)(x^3+1) x^3-2=0 x1=cbrt2, so we can do (x^3-2)/(x-cbrt2)= x^2+xcbrt2+cbrt4 to find the others. [1 quadratic formula later] x=+/-sqrt(-cbrt2-cbrt4) =+/-(isqrt(cbrt2+cbrt4)) x^3+1=0 x1=-1, do the same thing, (x^3+1)/(x+1)=x^2-x+1 [1 quadratic formula later] x=(1+/-isqrt3)/2 x1=cbrt2 x2=isqrt(cbrt2+cbrt4) x3=-isqrt(cbrt2+cbrt4) x4=-1 x5=(isqrt3+1)/2 x6=(1-isqrt3)/2
P2-P-2=0 we use quadratic trinomial decomposition A x B = -2 and A+B = -1 .......(P-2)(P+1)....in order for these 2 numbers to "come out", both are negative, one term must be + and the other minus, and since it is a simple quadratic equation, there must be 1 and 2😉 -2x1=-2 and -2+1=-1 this way is usually faster
If you taking an exam with time limit you wont make it. You already have n^3=2 take the cuberoot of 2 which is 2^1/3 thats the final answer substitute it to the original equation n^6-n^2=2 and your done no need to go to difference of two cube equation GOT IT!!!
@@onlineMathsTV, Sir today when I saw your this video again, after 5 months, i realised that you explained it so very well in detail.. When we have to find all the values including the imaginary once.. your way is the best.. Thank you so much Sir.. Regards 🙏🙏
@@josemaholo8651, we are taught this way for solving such problems mentally and checking the answers, before solving it.. this makes the approach easy and confident.. 😊🙏
My solution: N^6 = (N³)² Trick - N³ = X 1) X² -X -2 = zero By baskara, X' = 2 ; X" = -1 Then , N³ = 2 N = 1,25999 ( bY using my cell calculator ) Saludo from Brazil!!!
solved within 5 mins, and that's because I have not touched any math question for over a decade, that's junior highschool grade math, all you need to know is (a+b)^2 = a^2 + 2ab+ b^2,
Base on my algebra knowledge you can not do 2 unknown value cause i have examples too x + x = 10 5+5 is 10, 6+4 is also 10, 7+3 is also 10 but it is impossible to pick in this 3 solution because they are two unknown value
n^6-n^3-2=0 n^3=(1+-sqrt(1+8))/2=1+-3=4 or -2 n=cbrt(4) or cbrt(-2) RRT would have you believe there are 6 roots, so I’ll do it the cubic way: n^3(n^3-1)=2 n^3(n-1)(n^2+n+1)=2 I guess this would find them but i don’t see how to get rid of the 2.
By International math Olympiad, does it mean IMO? Cuz i thought IMO questions would be very hard but this was taught in my school and i am not particularly good at math anways.
Hello sir. Once we get to n^3= 2 or n^3= -1 (5 min into the video), isn't the problem already almost solved? I mean, raise both equations to the power of (1/3) and you get, right away, the solutions, namely n= 2^(1/3) and n = (-1)^(1/3) = -1, and it's done. These are the only 2 real solutions of the initial equation; I mean, n^3 is a continuous function whose derivative is always positive (other than for n=0) and so there is only one solution for each n^3, whatever n^3 is. Of course we may try to work out the other solutions (which is what you did), but we already know that the other solutions won't be real but rather imaginary solutions.
You are very correct sir. The further work is just to prove that the other solutions are imaginary. Thanks for watching our content and dropping this wonderful observation sir. Much love from all of us @onlinemathsTV ❤️❤️💖💖💕💕💕
even exponent makes negative number positive odd exponent makes negative number negative its definitely and clearly not 2 because 2^n is very famous (binary numbers) Solve these problems without algebra. Use theories, theorems, and common sense if you plug in values like 2 its clearly not 2 so it must be under 2 try -1 and 1 1 gives it 0 -1 gives it 2
once you get to P=n^3 = 2, you can go straight to n=2^1/3, you don't need to do all that other crap. Cube roots of real numbers are evenly spaced 120 degrees apart on a circle in the complex plane, with radius = real cube root. but there's only 1 real root in the X axis.
It is not a triangle per say...It is the symbol for discriminant which is used to check if a quadratic equation will give real or imaginary roots instead solving the the full quadratic equation using the formula method. Thanks for asking sir. Much love.....💕💕
Call n^3 = y and you get y^2 -y -2 = 0 , a second degree equation. ... y1 = 2 and y2 = -1.... So you Will have: n1 = 2^(1/3) and n2 = -1 Who is wrong ? Me or you ? In fact (2^(1/3))^6 = 2^2 = 4 And..... (2^(1/3))^3 = 2 And 4 - 2 = 2 (ok) For otherwise (-1)^6 - (-1)^3 = 1 - (-1) = 2 (ok) Who is correct ? Me or you ???
I am 66 but love maths and didn’t love it enough in high school! Thanks.
Thanks for the tips to determine if a quadratic function will give real roots or not
You welcome always...much love 💖💖💖
There are two real solutions. Substitute x=n^3 and you get the quadratic x^2 -x = 2. Then you will find that n = -1 and n= 2^(1/3)
Very very correct @Tony Martinez. You are good at it sir.
Much respect and love to you boss...👍👍💖💖
Exactly
❤
&G?{❤😊😂
8
That was my thinking too
Thank you for reminding us that there are, in general, six solutions to a 6th order equation. It would be more challenging to ask for all of the distinct solutions. Excellent video!.
100% agree
تمرين جميل جيد . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
Wonderful
Wow!!! what a wonderful prayer from a wonderful human/fellow from Gaza, Palestine.
A loudest amen I and my team say to your awesome prayers sir.
All of us @onlinemathstv love you and we also pray the Almighty God to protect and shine His face on you and your entire family members.
Hope to meet with you one-on-one some days to come.
We love you sir....💖💖💕💕❤️❤️😍😍
Thank you sir now I'm understanding the concept respect🙏🙏🙏
step 2:45 to 4:00 normally done in one go. thanks for the vid
Great explanation 👌
More interesting to cover all 6 solutions, and their intertwined relationship with the complex roots w where w^3 = +1
Excellent. But in Japan , Highschool students solve this question with complex plane.
if n^3=2 , they think that n = r(cos𝜽 + isin𝜽) then other solution are 2^1/3(cos2/3𝛑+isin2/3𝛑) and 2^1/3(cos4/3𝛑+isin4/3𝛑), I think.All solution in this case exist 6 pattern.
√-3=√-1×√3の式変形について、今回は良かったけど基本的にはa.b>0の条件付きで√a×√b=√abが成り立つので、数学的には良くないことをしてる事を自覚してから動画出していただきたい。
Thank you. a better and shorter solution is (N^3) (N^3 - 1) = 2 and since 2 = 1x 2, then (N^3 - 1) = 1 and N^3 = 2
Yes, but only applicable in case of request of integer solutions. In the case of request of "real" values, prime decomposition is irrelevant.
It's could be easier to do this with quadratic function and delta but this answer is also good
Well explained. God bless you too.
It' s much simpler (just a couple of passages...) using Ruffini' s rule, once you recognize that n = -1 is one of the roots. The other real solution is cubic root of 2 (2^(1/3))
Bravo....👍👍👍
Literally,You are legend 😮😮❤❤❤ keep it up
🇮🇶🇮🇶🇮🇶Good explanation. The topic of mathematical foundations requires multiple skills and ideas. From Iraq
Smiles...thanks for the compliments and thanks for watching our contents.
Much love from Onlinemathstv 💕💕❤️❤️💖💖
We can get six solutions .i give you three.
n= -1 or n=(2)^1/3 or n=(i)^2/3
.....
Thank you bro.
Thank you.this is very useful
n^6-n^3=2
-2 from both sides
let x=n^3
x^2-x-2=0
(one quadratic formula later)
x=2
x=-1
x=n^3
=> n=cbrt(-1)=-1
n=cbrt2
edit: ima try to find the imaginary roots just for fun
x^6-x^3-2=0, if y=x^3, y^2-y-2=0
to factor y^2-y-2, 2 numbers (a & b) must satisfy a+b=-1 & ab=-2
=>a=-2 & b=1
=> y^2-y-2=(y-2)(y+1)
y=x^3, => x^6-x^3-2=(x^3-2)(x^3+1)
x^3-2=0
x1=cbrt2, so we can do (x^3-2)/(x-cbrt2)= x^2+xcbrt2+cbrt4 to find the others.
[1 quadratic formula later]
x=+/-sqrt(-cbrt2-cbrt4)
=+/-(isqrt(cbrt2+cbrt4))
x^3+1=0
x1=-1, do the same thing, (x^3+1)/(x+1)=x^2-x+1
[1 quadratic formula later]
x=(1+/-isqrt3)/2
x1=cbrt2
x2=isqrt(cbrt2+cbrt4)
x3=-isqrt(cbrt2+cbrt4)
x4=-1
x5=(isqrt3+1)/2
x6=(1-isqrt3)/2
You the best sir....😍😍😍
P2-P-2=0 we use quadratic trinomial decomposition A x B = -2 and A+B = -1 .......(P-2)(P+1)....in order for these 2 numbers to "come out", both are negative, one term must be + and the other minus, and since it is a simple quadratic equation, there must be 1 and 2😉 -2x1=-2 and -2+1=-1 this way is usually faster
Great approach boss.
Lovely sir.
You the best because you are good at what you know.
Respect sir...👍👍👍
very good my friend
Great job friend🎉 I am Indian
If you taking an exam with time limit you wont make it. You already have n^3=2 take the cuberoot of 2 which is 2^1/3 thats the final answer substitute it to the original equation n^6-n^2=2 and your done no need to go to difference of two cube equation GOT IT!!!
Got it master. Thanks for this approach, you the boss here.
Respect 🙏🙏🙏
Great👍.
Sir we can also do it by hit and trial method, that is by assuming n=0, 1, -1, 2, -2..... We can get the answer quickly 🙏.
Regards
Very very correct sir.
Respect.....👍👍👍
Too long procedure for nothing...
@@onlineMathsTV, Sir today when I saw your this video again, after 5 months, i realised that you explained it so very well in detail..
When we have to find all the values including the imaginary once.. your way is the best..
Thank you so much Sir..
Regards 🙏🙏
@@josemaholo8651, we are taught this way for solving such problems mentally and checking the answers, before solving it.. this makes the approach easy and confident..
😊🙏
clearly communicated and easy to follow, great work!
Glad it was helpful!
Thanks for dropping a comment to encourage us sir.
Much love💕💕💕
"8*T>`&q)L )5Ps\ U wpQ"aP
My solution:
N^6 = (N³)²
Trick - N³ = X
1) X² -X -2 = zero
By baskara,
X' = 2 ; X" = -1
Then , N³ = 2
N = 1,25999 ( bY using my cell calculator )
Saludo from Brazil!!!
Excelente aula !
Thanks for appreciating our little effort sir.
Also thanks for watching and dropping this wonderful words of encouragement.
Excellent Sir
great teacher
Reasonless effort
🌈🌈🌈Rainbows
Excelent video.
Thanks for watching and leaving behind this wonderful comment sir. We love you....💕💕💕😍😍😍
good exercise thank you
You are most welcome sir.
We are glad you gained some values from our video.
Much love sir...💖💖💕💕
решается в уме.
пусть n^3=a.
т.е. у нас а^2-а=2
а(а-1)=2*1
следовательно а=2
n^3=2
n=2^(1/3)
При а=-1
n =-1
Bravo👍👍👍
excellent video
so cool :) have a nice day!
Thanks a million sir. You have a wonderful and a lovely day sir.
Much respect my friend...🙏🙏🙏👍👍👍👍.
Задачка тянет максимум на уровень проверочной работы восьмого класса
solved within 5 mins, and that's because I have not touched any math question for over a decade, that's junior highschool grade math, all you need to know is (a+b)^2 = a^2 + 2ab+ b^2,
Was thinking the same, this is not even close to be an "olympiad question"
Very beautiful
Voce explica muito bem
Deixa a matemática fácil 😘
Smiles, thanks for this wonderful encouraging words of urs.
@@onlineMathsTV 🙏
Nice job, as usual.
Una maniera assurda per complicare un calcolo semplicissimo che si può fare a mente... Ridicolo!
This is solved in the mind
sure
Muy bien hermano.
Thanks a million my dear.
Much love 💖💖💕💕😍
Thanks
You are most welcome sir. One love from all of us @onlinemathstv.
Можно через замену n^3=t, t^2-t-2=0, t=-1,t=2, n= -1, n= корень кубический 2.
Да! Это очень простая задача. Если это реально есть в чьей-то олимпиаде, то мне жаль их систему образования.
65=81 - 16
or 81=3exp 4 and 16=2exp 4
So m=4 unic solulution
Really, this was a question in IMO? The questions I got in INMO( Indian National Mathematics Olympiad) were much harder than this.
数学って素晴らしいな
喋る言語は違うけど、繋がれる。
nice job
thanks for the appreciation
Great job, professor!
Which phone are you using to do this pls? And light system?
n^6-n^3= 2
ada 6 solusi untuk n,
Jika
n^3 = a
Maka, a^2-a= 2
a^2 - a - 2 = 0
(a-2)(a+1)=0
a=2 atau a=-1
Jadi
n^3= 2 atau n^3= -1
Jika sumbu Y adalah imaginer
Dan sumbu X adalah Real
Maka:
n^3= 2
n1= 2^(1/3) * [Cos 0° - i*Sin 0°]
n2= 2^(1/3) * [Cos 120° - i*Sin 120°]
n3= 2^(1/3) * [Cos 240° - i*Sin 240°]
n^3= -1
n4= [Cos 60° - i*Sin 60°]
n5= [Cos 180° - i*Sin 180°]
n6= [Cos 300° - i*Sin 300°]
Disederhanakan maka:
n1= 2^(1/3)
n2= -2^(1/3) * [i*(√3)+1]/2
n3= 2^(1/3) * [i*(√3)-1]/2
n4= [1-i*(√3)]/2
n5= -1
n6= [1+i*(√3)]/2
Jadi ada 2 solusi nilai n= real dan 4 solusi nilai n= imaginer
Prof, you are good at what you do sir.
Respect and thanks for dropping this procedure.
Base on my algebra knowledge you can not do 2 unknown value cause i have examples too x + x = 10 5+5 is 10, 6+4 is also 10, 7+3 is also 10 but it is impossible to pick in this 3 solution because they are two unknown value
Thanks for a job well done Jakes
You are welcome all the time ma
Great job at that
n^6-n^3-2=0
n^3=(1+-sqrt(1+8))/2=1+-3=4 or -2
n=cbrt(4) or cbrt(-2)
RRT would have you believe there are 6 roots, so I’ll do it the cubic way:
n^3(n^3-1)=2
n^3(n-1)(n^2+n+1)=2
I guess this would find them but i don’t see how to get rid of the 2.
Just spotted n=-1 as a solution
Nice
Thanks sir...❤️❤️
By International math Olympiad, does it mean IMO? Cuz i thought IMO questions would be very hard but this was taught in my school and i am not particularly good at math anways.
lmao this is so easy, i did in like 20 seconds, i dont think it would be something that pops up on the IMO 😂
yeah fr the title is clickbait
It’s -1, right? (-1) to the even power is 1, and (-1) to the odd power is -1, so 1 - (-1), or 1 + 1 = 2.
i found six rooths
n = {-1 , (1+isqrt(3))/2 , (1-isqrt(3))/2 , cbrt(2) , -cbrt(2) +- isqrt(3cbrt(2)) }
👍👍👍 you the guru @Lucien34.
@@onlineMathsTV guru ?
Well depth in the subject matter… sorry for the jargon
Why did I get results different than yours (root #1: n = ³√2 and root #2: n = -1)? See below:
Question:
n⁶ - n³ = 2
n = ?
Answer:
n⁶ - n³ = 2
if N = n³ then n⁶ - n³ = 2 = N² - N = 2
N² - N - 2 = 0
delta = (-1)² - 4*1*-2 = 1 - (-8) = 9
√delta = √9 = 3
root #1:
N = (-(-1) + 3)/2*1 = 4/2 = 2
N = n³ = 2 => n = ³√2
root #2:
N = (-(-1) - 3)/2*1 = -2/2 = -1
N = n³ = -1 => n = ³√(-1) = -1
Result(s):
n = ³√2
n = -1
🙂
I will go through mine and urs and get back to you sir. Just give me some time.....
Вы получили на 100% идентичный вариант. Другое написание
GENIAL!!!
Faurmidable!Excellent
Thanks a million sir.
Merci
I found the roots for n^2 - n + 1 = 0 using the quadratic equation to be [SQR (3) i + 1]/2 and [SQR (3) i -1]/2. Why did you use only delta?
It is used just to know if the quadratic equation will give imaginary or real roots. It is a formula know as the "discriminant factor".
Nice math
Thanks sir.
We love you for watching and comment sir.....💖💖💖
-1 is one answer. 1 - (-1) = 1 + 1 = 2. The other answer is 2^(1/3). 4 - 2 = 2.
Very correct ma.
Bravo 👍👍👍
Respect sir 🙏🙏🙏
Hello sir.
Once we get to n^3= 2 or n^3= -1 (5 min into the video), isn't the problem already almost solved? I mean, raise both equations to the power of (1/3) and you get, right away, the solutions, namely n= 2^(1/3) and n = (-1)^(1/3) = -1, and it's done. These are the only 2 real solutions of the initial equation; I mean, n^3 is a continuous function whose derivative is always positive (other than for n=0) and so there is only one solution for each n^3, whatever n^3 is.
Of course we may try to work out the other solutions (which is what you did), but we already know that the other solutions won't be real but rather imaginary solutions.
You are very correct sir. The further work is just to prove that the other solutions are imaginary.
Thanks for watching our content and dropping this wonderful observation sir.
Much love from all of us @onlinemathsTV ❤️❤️💖💖💕💕💕
even exponent makes negative number positive
odd exponent makes negative number negative
its definitely and clearly not 2 because 2^n is very famous
(binary numbers)
Solve these problems without algebra. Use theories, theorems, and common sense
if you plug in values like 2 its clearly not 2 so it must be under 2
try -1 and 1
1 gives it 0
-1 gives it 2
Parabéns irmão. Ótimo trabalho!
👍👍👍
Thanks for the encouragement sir.
We love you for watching and at the same time commenting.
Use log
n^6 - n^3 = 2
log n^6 - log n^3 = log 2
6 log n - 3 log n = log 2
3 log n = log 2
log n = 1/3 log 2
log n = log 2^1/3
n = 2^1/3
When you have n^3=2. You should have cube root on both sides so n=3√2
ok
Raíz cúbica de 2 también es valor de n
Yes, you are very correct sir.
-1 or 2^(1/3)
Giải bằng nhân tử hoặc giải bằng phương trình bậc 2 theo delta
That's another wonderful way out too.
once you get to P=n^3 = 2, you can go straight to n=2^1/3, you don't need to do all that other crap. Cube roots of real numbers are evenly spaced 120 degrees apart on a circle in the complex plane, with radius = real cube root. but there's only 1 real root in the X axis.
Congratulations, a black man who is not showing off with football, dancing, but with exemplary disciplinary content.
😂😂😍😍
Thanks for this wonderful comment sir.
Respect
Đúng là giải toán kiểu tây. Dài dòng. Học sinh VN giải trong phút mốt.
Wow, nice and great brains.
You can use log it will help
Valeu professor!!!
I didnt get what you did at the ending when you introduced the triangle...you didnt use the full quadratic formula
It is not a triangle per say...It is the symbol for discriminant which is used to check if a quadratic equation will give real or imaginary roots instead solving the the full quadratic equation using the formula method.
Thanks for asking sir.
Much love.....💕💕
@@onlineMathsTV you're the best sir 🥰
Pls help me to understand why any number exponent o is always 1
Al ojo: raíz cúbica de 2
Why you don’t use Log it will be very simple
Very long slution but good
for me the Answers are cube root of 2 and -1
is it just me or can you not just cube root 2 and -1 after you find the quadratic
p1+p2=1. p1×p2=-2. p1=2. p2=-1
tambem podia se "raiz cúbica de 2".
x^2 -x=2
x^2-x-2=0
x=0.5*(1±√(1-(4×1×-2)))
x=0.5*(1±√(1+8))=0.5(1±√9)
x=0.5*(1±3)=0.5*(4, -2)
x=2, -1
x=n^3 = 2, -1
If complex ones, just multiply (1+-i sqrt(3))/2 on both p
Noted sir
2 reality and 4 complexity
Sir, this kind of calculation musst always ends with check! School 8 grade. Third root of 2 is a wrong solution.
sorry I didn't state it
Is this considered the easiest question in Olympiad?
No offense, just curious
Solution:
n⁶ - n³ = 2
let x = n³;
x² - x = 2 ⇒ x² - x - 2 ⇒ (1 ± √9)/2 ⇒
x₁ = 2; x₂ = -1
Restore x = n³;
n₁ = ³√2
n₂ = ³√(-1) = -1
1--1=2 n= -1
Bro you can just use cube root directly
Call n^3 = y and you get y^2 -y -2 = 0 , a second degree equation. ...
y1 = 2 and y2 = -1.... So you Will have:
n1 = 2^(1/3) and n2 = -1
Who is wrong ? Me or you ?
In fact (2^(1/3))^6 = 2^2 = 4
And..... (2^(1/3))^3 = 2
And 4 - 2 = 2 (ok)
For otherwise (-1)^6 - (-1)^3 = 1 - (-1) = 2 (ok)
Who is correct ? Me or you ???
Both of us are correct......😍😍😍😍😍
Thanks for your approach and comment sir.
n=exp(i*Pi) j'ai trouvé en 10 secondes ...