A Polynomial Equation Adapted from Putnam Exam 2006

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#ChallengingMathProblems #Putnam #polynomialequations
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[@SyberMath]

SAME PROBLEMS FROM DIFFERENT PERSPECTIVES Solving A Cubic Equation in Two Variables: kzbin.info/www/bejne/l363g6aobsiAbNk Factoring a cubic polynomial: kzbin.info/www/bejne/eYGof3qLjsSWo7M

[@mohammadamir1985]

Saw this on Michael Penn's channel. But problem was phrased little differently. This equation shows a glimpse in elliptic curves. Beautiful😍💓. Good place to stop🤣

[@SyberMath]

Thank you!

[@willnewman9783]

I don't think the equation relates much to elliptic curves. Yes it is a cubic equation, but it is singular, so its solution set is not an elliptic curve.

[@idontknowwhatido3972]

Great Mohammad, I would love if your muslim people taught algebra again as in the middle ages! I miss it so much, I try to study muslim scholars in math in the middle ages. Make it come again please :)

[@roberttelarket4934]

Good place to start some more SyberMath problems.

[@shmuelzehavi4940]

@@willnewman9783 You are right.

[@aashsyed1277]

Cool syber! Very! I hope this attracts views so that you get more subscribers!

[@aashsyed1277]

Cool video! Thanks so much for these joyful moments ❤️💕

[@SyberMath]

💖

[@maryrose9385]

All real sols for this equation are (x,y)=(-1,-1);(t,1-t) where t is a real number

[@carloshuertas4734]

Another great explanation, SyberMath! I thought that it was just one solution. I figured it out. Thanks!

[@SyberMath]

You're welcome!

[@sea34101]

I found a different proof: Step 1 adding and removing 3x.x.y + 3.x.y.y we end up with (x+y)^3-1 that we can factor as a difference of cube, and we get the formula at 3:50 Then, to factor x.x + y.y + 1 - x.y + x+y, we can do the change of variable x = u+v and y = u-v and some square expressions appear naturally and there is one single solution: u=-1 and v=0 hence x=y=-1 Hence we have all the solutions ^^

[@chessdev5320]

we have to use the sum of three cubes identity here and then its simply equating it with 0. It can be factorised by x³+3xy+y³-1 = (x+y-1)(x²+y²+1-xy+x+y)= 1/2(x+y-1)((x+1)²+(y+1)²+(x-y)²)= 0 hence, i think it has infinite real solutions of x and y which satisfy x+y=1 Or x=y {which has (-1,-1) and (1/2,1/2) as solutions}.

[@arpansit3155]

This is just a amazing method to aprroach at the solutions. Awesome sir. Congo for ur 30k subscriber.

[@SyberMath]

Thanks a ton! 💖

[@rudranarayanswain2382]

This approach was amazing..never thought of that...👍

[@mustafizrahman2822]

Thanks for your nice explanation.

[@SyberMath]

You are welcome

[@manojsurya1005]

Dude the factoring method was awesome 🤩

[@SyberMath]

Glad you liked it!

[@MizardXYT]

Because the equation is symmetrical, I like the substitution { x = u + v, y = u - v } This turns x² + y² - xy + x + y + 1 = 0 into u² + 2u + 1 + 3v² = 0 which can be simplified as (u + 1)² + 3v² = 0. The only solution is { u = -1, v = 0 }. Substitution back to x and y gives { u = (x + y)/2 = -1, v = (x - y)/2 = 0 } witch is solved by { x = -1, y = -1 }

[@danielwest4861]

When I worked on it first, I noticed it would look identical to the given equation if I set x+y equal to 1. I’m not sure if my approach was a legitimate solution but that’s how I did it. (x+y)^3 = x^3 + y^3 +3xy(x+y) Love your videos as usual!

[@naharmath]

we have: x^3+y^3=(x+y)^3-3xy(x+y). Then the equation becomes: (x+y)^3-1-3xy(x+y-1). By using the equality (x+y)^3-1=(x+y-1)((x+y)^2+(x+y)+1) we obtain the needed factorisation. After that we have a simple equation x^2+(1-y)+y^2+y+1=0. Its discriminant is delta=-3(y+1)^2

[@asaradhi]

Fantastic approach in using the identity

[@SyberMath]

Glad you think so! 💖

[@kaslircribs5804]

Excellent! Thanks professor!

[@SyberMath]

You're very welcome!

[@kshitijjain9302]

Good one, I knew this question could be solved in this way

[@SyberMath]

Nice!

[@MathElite]

Another awesome problem on Syber's channel I got a solution: x+y=1

[@SyberMath]

Thank you, Math Elite! 💖

[@joaquingutierrez3072]

Amazing video!

[@SyberMath]

Glad you think so!

[@diogenissiganos5036]

That z sub was absolutely ingenious

[@SyberMath]

Thank you!

[@satyapalsingh4429]

You are genius .

[@vacuumcarexpo]

I couldn't find (x－y)^2+(x+1)^2+(y+1)^2＝0. On second thought, this is the same transformation used to prove the inequality of arithmetic and geometric means of 3 terms. I should have been aware of this. Instead of that, I used (x+1)^2+(y+1)^2－(x+1)(y+1)＝0. Let X＝x+1 and Y＝y+1, X^2－YX+Y^2＝0 X,Y∈R⇒D＝Y^2－4Y^2＝－3Y^2≧0⇒Y＝0⇔y＝－1⇒x＝－1 ∴(x,y)＝(－1,－1) I solved the latter half of this question like this.

[@SyberMath]

Nice!

[@user-sl9mx5gc4s]

From my observation, one solution is x=y = -1. This solution satisfies the equation. It seems you don’t need to any work. Pls let me know if I made a mistake. Thx.

[@zakirreshi6737]

Dividing by y^3 and putting x/y =u

[@tonyhaddad1394]

Awesome approach , your great 😍

[@SyberMath]

Thank you! 😃

[@idontknowwhatido3972]

"You are" or "You´re", people learn it finally. I can remeber it although my english is bad and I am an imbecil, so you can learn it too!

[@thomasnielsen1187]

10:42 And the solution is a line where y=-x+1 :-o Great equation. Great method. Great solution. And a great video! Yes he found that solution 5 min. before. But It was only the 2nd time it was written that I realized it... :-D

[@SyberMath]

Thank you! 💖

[@abuobidashihab]

satisfying solution ♥️

[@SyberMath]

Thanks!

[@willyh.r.1216]

Brilliant.

[@SyberMath]

Thank you!

[@kyriakosmonodri4601]

Maybe because the expression is symmetrical, should x=ψ?

[@수하긴]

Lovely

[@SyberMath]

Thanks!

[@davidbrisbane7206]

You'd think from the symmetry of the equation that one of the solutions must be x = y. So, x^3 + 3xy + y^3 = 1 with x = y Then 2x^3 + 3x^2 = 1 So, 2x^3 + 3x^2 - 1 = 0 So by inspection, x = -1 is a solution. So, x = y = -1 is a solution.

[@SyberMath]

Nice!

[@albertmcchan]

No. symmetry only gives that if (x,y) = (x1,y1) is solution, then (x,y) = (y1,x1) also satisfy. This is for the same as x1=y1.

[@broytingaravsol]

at least x=-1, y=-1

[@alessandrorossi1294]

I took the 2006 Putnam exam. I don't remember if I got a point or not

[@SyberMath]

Wow! Even taking the test would be a huge success! 😍

[@tmacchant]

The discriminant D=-3(y+1)^2. If y=-1, D=0, otherwise D

[@rssl5500]

Nice

[@SyberMath]

Thanks

[@jimmyzhang8443]

It is a super action when you X2 at 8minutes

[@SyberMath]

I agree! 😊

[@petereziagor4604]

I did enjoy it but you've got to make things clearer next

[@SyberMath]

How?

[@idontknowwhatido3972]

No need friend, it was all crystal clear. Please learn more yourself, lacking in knowledge is on your part. Peace friend.

[@petereziagor4604]

@@idontknowwhatido3972 okay Maybe my class shouldn't be learning maths like this

[@OkiDoki.a]

Cool

[@adgf1x]

X+y=1

[@SyberMath]

Why?

[@leecherlarry]

my compi is confused haha: *Reduce[x^3 + 3 x y + y^3 == 1, {x, y}, Reals] // FullSimplify // TraditionalForm*

[@MathElite]

rip compi

[@charly6052]

Programation

[@leecherlarry]

@Nirupam dn lol guys . compi did :D

[@SyberMath]

Yessss! 😁

[@Germankacyhay]

👍❤

[@SyberMath]

😊

[@xuantri334]

Hay

[@뚜비-l8m]

ㆍ6

[@pisethmath4576]

help me sinx=x/100

[@SyberMath]

kzbin.info/www/bejne/oXK6eoali5eHaas