Elegant solution. You are a great educator. 👍

@PreMath

2 жыл бұрын

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Geometer. Keep it up 😀 Love and prayers from the USA!

There is also a graphical proof for the result: The x value of (-1+i*sqrt(3))/2 represents a vector on the unit circle on the complex plane which shows in the eleven o'clock position or 120° rotated counterclockwise from the x^0=1 (vector directed on the three o'clock position). So x^(n+1)=x^n*x actually rotates the vector 120° counterclockwise. And on x^3, you get the x^0=1 unit vector back. So all x^(3*n) represent unit vectors to the three o'clock position, x^(3*n+1) represent unit vectors on the eleven o'clock position, an x^(3*n+2) are directed to the seven o'clock position. So on the term x^77+x^76+x^75+x^74+x^73 we just have to caculate the 3 modulus of the exponents: 77 mod 3 = 2 => 7 o'clock unit vector, 76 mod 3 = 1 -> 11 o'clock vector, 75 mod 3 = 0 -> 3 o'clock vector, 74 mod 3 = 2, 73 mod 3 = 1. Since we get 0 once and 1 and 2 twice, we walk one unit to the right (=+1) but 4 times a half step right (both x^1 and x^2 have -1/2 as real part). So +1-4*(1/2) = -1 results. On the imaginary axis we walk sqrt(3)/2 up twice (n mod 3=1) and also sqrt(3)/2 down twice (n mod 3=2) which cancels out so the imaginary part stays 0. => This results to -1+0*i = -1 as shown in your video. Drawing a geometric figure for that is very easy for that.

The steps towards the solution is so inspiring.

really strategic and incredible explanation!🎉 Hope to see ones like that in new videos❤

Amazing method prof.great job

One of my favorite maths solutions

Actually the most important thing is to prove x^3=1. After that x^77=x^2, x^76=x,.... Hence, the requested polynomial becomes x^2+x+1+x^2+x=2(x^2+x+1)-1=-1

@MichaelRothwell1

2 жыл бұрын

That's the way I did it. Nevertheless, the solution presented was quite elegant. My only criticism was the way the first method was so readily abandoned. If the Cartesian solutions had been converted to polar/Euler form, it have been a piece of cake.

Since x^3 =1, x has three different values, two of which are complex conjugates of one another, the third being 1.

سؤال يااستاذ من اين اتينا بمعادلة فرق المكعبين؟ارجو التوضيح

Its actually a veryy easy peoblem if you know about cube root of unity ! : ) I solved it in my head...

Bravo!!

wow this was amazing

Simply x is the complex cube root of unity. Hence x^3=1. Hence x power any multiple of 3 is unity. This gives the answer as -1.

My approach: if x^2+x+1 =0, then x !=1, so you can multiply both sides by (x-1). Thus x^3-1=0 and x^3 = 1. If x^3 = 1, then x^72=1 and x^75 = 1. If (x^77+x^76+x^75+x^74+x^73) = y, then (x^77+ x^76+x^75) + (x^74+x^73+x^72) = y+1 y+1= (x^2+x+1)x^75 + (x^2+x+1)x^72 = 0 Thus our solution is -1.

@MichaelRothwell1

2 жыл бұрын

This is correct, but the stipulation that x≠1 to justify multiplying by x-1 is unnecessary. The reason is as follows: if you were trying to solve the equation, you would want to avoid adding extraneous roots, and so would need to add the caveat x≠1 (In such a situation you require successive statements to be equivalent). Here, however, we merely wish to deduce that if x²+x+1=0 then x³-1=0 (no equivalence required). It's rather like saying that if x=4 then x is even. The statements are not equivalent (they're not meant to be), but the second does follow from the first.

This is even better than my solution. For me, I went with solving for x to prove the x^3 = 1 part, and it was trickier than using that a^3 - b^3 formula, which stops you from having to get into complex numbers at all. The rest of what I did was the same as what you did.

@roger7341

8 ай бұрын

Not so difficult. x^2 = -x -1, x^3 = -x^2 - x = x + 1 - x = 1 and done.

Great!

What a great answer and the solution process itself! Thanks a lot, Sir! You are an excellent teacher indeed. My best wishing to you !

@PreMath

2 жыл бұрын

You are very welcome. Glad to hear that! Thank you for your feedback! Cheers! You are the best Anatol. Keep it up 😀 Love and prayers from the USA!

Sir another method to find x^3 :- while simplifying the given question at the step of x^73(x^3 + x^4) we can take x^2 common and we will get x^75(x^2 + x) which is again -x^75.,,so now -x^75 = -x^78 , we get x^3 =1

very well done bro, thanks for sharing

@PreMath

2 жыл бұрын

Thanks for the visit You are very welcome. You are awesome. Keep it up 😀 Love and prayers from the USA!

Very good

When x^3 =1, why we can not get x=1 here?

Thanks 👍

Ans : - x power 75

I came to X³=1 through another way. Just like the video, I've factored the expression until 3:47, which could be written as x^77 + x^76. But after that, I first tried factor out with: x^75(x^2 + x) Since x^2 + x + 1 = 0, x^2 + x = -1. So x^75(-1) = -x^75 Obviously I've came to a dead end, but seeing you factoring differently and since -x^75 and -x^78 came from the same expression, they must be equal. So I did: -x^78=-x^75 x^78=x^75 (x^75)(x^3)=x^75 x³=1 Finding this and plugging in back to -x^78 or -x^75 gave me the same answer.

Omega is one root to equation.w^3=1

This is a very easy problem for anyone who knows Complex Numbers

I have a little question sir ..... If we go on with x^3_1=0,then x^3=1,thus x=1...then the answer could be 5.But you just got _1 as the answer.where have i gone wrong?? Please explain sir...regards

If you get x^3=1, you also get x=1, why cant you just plug it in the original question directly? This would give a different solution though ...

Thanks you

Adding X^78 and subtracting the same value we get Y=x^78+x^77+x^76+x^75+x^74+x^73-x^78=x^76(x^2+x+1)+x^75(x^2+x+1)-x^78=-x^78. From that point I went into another direction (little bit harder way :-)) . The solution of the x^2+x+1=0 is in complex set: -(0.5)+/-i0.5sqrt(3). which can be presented as cos(2pi/3)+i*sin(2pi/3) and that is equivalent to e^(i2pi/3). [Let exclued 2*k*pi part for a moment] Now Y=-x^72=-e^(i48pi)=-(cos(48pi)+isin(48pi))=-1. Prof, I admire your approach because is simpler! The solution of x^2+x+1=0 is a complex solution of equation x^3=-1 too. Knowing that the solution of Y=-x^72 is easy...

@marioperic9709

2 жыл бұрын

I correct Y=-X^78 instead -x^72. But the same way is if we add x^72. the solution is the same because the 72 and 78 are even multipliers of 3.

@geometer6121

2 жыл бұрын

Outstanding! 👍

Nice

I was able to figure out a shortcut to the solution. First I mulitiplied the equation with x⁷³, I ended with : X⁷⁵ + x⁷⁴ = -x⁷³ Then I plugged in the value of x⁷⁵ + x ⁷⁴ in the question and they canceled with the x⁷³. Now I was only left with x⁷⁷ + x⁷⁶. The rest is the same with your solution. I found the value of x³ and plugged it in.

yes!!!

By multiplying the given expression both sides by x-1, we get x^3-1 =0 . So x=1 Has it any loophole?

@aryan4679

2 жыл бұрын

Yes man i am confused why he got -1 , x =1 only

@stevehernandez2510

2 жыл бұрын

x never equaled 1. he multiplied by x-1 to get an easier polynomial to work with. It's the same as finding multiples of 25 by multiplying 25 by 4 because using 100 is just easier to deal with

@aryan4679

2 жыл бұрын

@@stevehernandez2510 sorry sir, but i am still confused

Ans : 0 I guess?

It's just omega...w^3n=1,w²+w=-1

A fine problem, but the non-integer result is slightly disappointing. If 84 were replaced with 156, and 35 replaced with 65, then all line-segments in the figure would have whole-number lengths, and the proportions would be exactly the same. If, instead, the diagonal lengths were 20 and 15, then all line-segments would again be integers, based on scaled 3-4-5 triangles. I think the minimum lengths for scaled 7-24-25 triangles would be 600 and 175. A simple formula is hiding there somewhere.

V nice approach vnicely solved

@PreMath

2 жыл бұрын

Keep watching Thank you for your feedback! Cheers! You are awesome Niru. Keep it up 😀

easy in tenth standard i was able to do it in less than a min just done reduction

Very elegant - thank you!

@PreMath

2 жыл бұрын

You are very welcome. Glad to hear that! Thank you for your feedback! Cheers! You are awesome David. Keep it up 😀

Amazing way of teaching👍 Thanks for sharing😊

@PreMath

2 жыл бұрын

You are very welcome. Glad to hear that! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

Correct

@PreMath

2 жыл бұрын

Thank you for your feedback! Cheers! You are awesome Abasher. Keep it up 😀

I WILL SUGGUST TO BE USE CIS FIRST LET X^2+X+1=0 BOTH SIDE MUPLIT X-1 and you can get that will be x^3-1=0 so the soultion will be cis(pi/3) x^77+.....=x^73(x^4+x^3+x^2+x+1)=x^73(x^4+x^3)=x^73(x^4+x^3+x^2-x^2)=x^73(-x^2)=-(x)^75 -(x)^75=-cis(pi/3)^75=-cis(pi75/3) =-cis(pi25) =-(cis(pi))^25 o=-1 anyway i am not sure if it will get extraneous root or not in use this method

Very nice sir!!! Can you tell me the whiteboard that you are using? Thanks a lot!!!

@PreMath

2 жыл бұрын

Thanks dear We use Camtasia Techsmith utility with photoshop. Cheers!

Thanks for the explanation!

@PreMath

2 жыл бұрын

You are very welcome. Thank you for your feedback! Cheers! You are awesome Matteo. Keep it up 😀 Love and prayers from the USA!

Your result is incomplete. The equation x³=1 has 3 solutions in C. I have found 72 valid solutions in C for the original problem. 1 is only one of them.

Can I share your videos in my blog website?

@PreMath

2 жыл бұрын

Please send a request through our email: premathchannel@gmail.com Thank you

Hello, I’ve been to several math olympiads and they never put that easy stuff anywhere, if it is from a math olympiad, please tell me from which one it is, othervise, dont put “MATH OLYMPIAD QUESTION” in your thumbnails

There is a 2nd approach, similar to the solution presented. Let y = x^77 + x^76 + x^75 + x^74 + x^73 y + x^72 = x^77 + x^76 + x^75 + x^74 + x^73 + x^72 y + x^72 = x^75*(x^2 + x + 1) + x^72*(x^2 + x + 1) => y + x^72 = 0 => y = - x^72 Because x^3 - 1 = (x - 1)(x^2 + x + 1) => x^3 - 1 = 0 => x^3 = 1 Therefore, y = - (x^3)^24 => y = - (1)^24 = -1

@geometer6121

2 жыл бұрын

Nice.

@PreMath

2 жыл бұрын

Excellent Clement! Glad to hear that! Thank you for sharing! Cheers! You are awesome. Keep it up 😀

First off, your solution was fantastic- very nice. I did the expansion of x, and while it's nowhere near as nice as the given solution, it is hardly hectic. Calculate x^2 either with x times x or as -1-x and you get (-1+/- i sqrt(3))/2. x^3 is then x*x^2, which is a difference of squares and comes out to 1. This shows a repeating pattern of values of x, x with inverse of i term, and 1. As 75 is a multiple of 3, x^75 is 1 and the terms around it are simplified to 2x + 2x^2, which is quickly calculated to 1. That being said ... I wish I had done it your way instead. :)

@PreMath

2 жыл бұрын

Thank you for sharing! Cheers! You are awesome Jay. Keep it up 😀

Nice problem + nice approach big fan of sums and of u

@PreMath

2 жыл бұрын

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Moorthy dear. Keep it up 😀 Love and prayers from the USA!

I note the complex solutions of the equation are unrelevant with the purpose of the exercise. In other words: what have the two complex solutions of x^2+x+1=0 to do with the rest of the procedure?

@yardenfrank357

2 жыл бұрын

We know x isn't 1 (try to substitute x=1 into the original equation) . Therefore, for the equation x^3=1 to remain true, x must be a complex number. To be more exact, x^2+x+1 is never equal to 0 in the real number system.

@PreMath

2 жыл бұрын

Great observation. Thank you for sharing! Cheers! You are awesome Fabio. Keep it up 😀

You solved lengthy man . Take omega as root of this equation and cubeof omega is 1. Thanks for question

@PreMath

2 жыл бұрын

You are very welcome. Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

I did a similar approach but using De Moivre's theroem

@PreMath

2 жыл бұрын

Very smart! Glad to hear that! Thank you for your feedback! Cheers! You are awesome Jasim. Keep it up 😀 Love and prayers from the USA!

x=1

The explanation in the video is unnecessarily longwinded. Since x² + x + 1 = 0 implies x³ = 1 we can take all exponents in the expression to evaluate modulo 3, so we have (x² + x + 1) + (x² + x) = 0 + (−1) = −1

@stevehernandez2510

2 жыл бұрын

I was thinking about that too

3 second problem

Nice problem and nice explanation of solution

@PreMath

2 жыл бұрын

Keep watching Thank you for your feedback! Cheers! You are awesome Rahul. Keep it up 😀

good approach

@PreMath

2 жыл бұрын

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Nico. Keep it up 😀

Finding the value of x was unnecessary. 😉

Nice Video!

@PreMath

2 жыл бұрын

Glad you enjoyed it

Very Nica solution! I would prefer to have this Problem in the form: Prove, that x^77+....+x^73=-1, when x^2+x+1=0 This would be a hint how far i have to Go. Tanks for your wonderful Channel!

@PreMath

2 жыл бұрын

Very good! Thank you for sharing! Cheers! You are awesome Stephen. Keep it up 😀

Can u take problems from hall and knight higher algebra problems.i would be thankful to u sir 😊

@PreMath

2 жыл бұрын

You are very welcome. Great suggestion! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

@INDIANATHEIST526

2 жыл бұрын

@@PreMath tks.i hv this book and rarely solved 30% of problems.

Op

Superb

@PreMath

2 жыл бұрын

Thank you! Cheers! You are awesome. Keep it up 😀

Thank you

@PreMath

2 жыл бұрын

You are very welcome. Thank you for your feedback! Cheers! You are awesome Cozma. Keep it up 😀 Love and prayers from the USA!

Figuring that x=1 was the punch line.Nice method.

@PreMath

2 жыл бұрын

Glad it helped! Thank you for your feedback! Cheers! You are awesome John. Keep it up 😀

@MichaelRothwell1

2 жыл бұрын

You mean that x³=1 was the punch line, right?

@stevehernandez2510

2 жыл бұрын

more like multiplying by x-1 was the punchline

Thnku

@PreMath

2 жыл бұрын

You are very welcome Pranav. Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

Wow!

@PreMath

2 жыл бұрын

Glad to hear that! Thank you for your feedback! Cheers! You are awesome JL. Keep it up 😀

Ça va très bien aussi avec les n’ombres complexes, sous la forme exp

@PreMath

2 жыл бұрын

Merci pour votre avis! Acclamations! Tu es génial Dominique. Continuez comme ça 😀

Great

@PreMath

2 жыл бұрын

Thank you for your feedback! Cheers! You are awesome Pranay. Keep it up 😀

very easy one

@PreMath

2 жыл бұрын

Great

I did it rightly till -x^78 and then couldnt solve it further. Awsm question btw

@PreMath

2 жыл бұрын

Thank you for sharing! Cheers! You are awesome Sreehari. Keep it up 😀

👍

@PreMath

2 жыл бұрын

Thank you dear for your feedback! Cheers! You are awesome. Keep it up 😀

👌👌👌👌👌👌

@PreMath

2 жыл бұрын

Super

رائع جدا

@PreMath

2 жыл бұрын

شكرا لك على ملاحظاتك! هتافات! أنت رائع يا تيجاني. استمر في ذلك 😀 حب ودعاء من امريكا!

@tijanimaths6006

2 жыл бұрын

@@PreMath انت القدوى بالنسبة لي🙏🙏🙏

@PreMath

2 жыл бұрын

@@tijanimaths6006 شكرا لك اخي. أنت كريم جدا😀

x^2 + x +1 = 0 is to be multiplied by x. - 1, the expression becomes x^3 - 1= 0 x^3 = 1 Now the given expression can be written as x^75( x^2 + x +1) + x^72( x^2 + x + 1) - x^72 = 0+0-x^72 = -(x^3)^24 = -1 ans

@mohanvaddadi

2 жыл бұрын

If x^3 is 1, is not x=1 , Any power of 1 is 1 and so answer is 5. Please explain where I went wrong.

Damn wow

@PreMath

2 жыл бұрын

Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

Solved sir

@PreMath

2 жыл бұрын

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are awesome Kumar. Keep it up 😀

New approach !!!

@PreMath

2 жыл бұрын

Yes, it is! Thank you for your feedback! Cheers! You are awesome Govinda. Keep it up 😀

If I substitute negative 1 in the polynomial I wouldn't get zero!!!

I think the answer will -1. Ok, the answer is -1.

@PreMath

2 жыл бұрын

Well done Mustafiz!

@rahulpaul5539

2 жыл бұрын

Yeah

@mustafizrahman2822

2 жыл бұрын

@@rahulpaul5539, Can you tell me why the Indian questions ( NEET, JEE, and others) are so hard? What's wrong with you?

@mustafizrahman2822

2 жыл бұрын

@@rahulpaul5539 Can you tell me how to be well prepared in JEE Advanced?

@mustafizrahman2822

2 жыл бұрын

@@rahulpaul5539 As a Bangladeshi, I wanna read like Indian JEE Chanced level student or mathematician, physics and chemistry lover students. I think most of them get into JEE Advanced. Please please suggest me

x=yi __ i=-,/-1 y=?

@olgakornak3969

2 жыл бұрын

x=1?

First view

@PreMath

2 жыл бұрын

Keep watching Thank you for your feedback! Cheers! You are awesome Amav. Keep it up 😀

Solved sir

@PreMath

2 жыл бұрын

Super Vats You are awesome. Keep it up 😀