A Nice Polynomial System

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#ChallengingMathProblems #PolynomialSystems

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[yoav]

I come evry day just to hear" helo every one", the rest is bonus.

[SyberMath]

Hello everyone!
I uploaded the thumbnail like 12 hours ago but I noticed it was not showing a little before the video premiered! I don't know what it is but it seems be a YouTube glitch.

EDIT: The thumbnail shows up now. Thank you @Math Elite for letting me know! Btw, check out his channel here:

[BharatSharma-ofdu]

I counted your subscribers exactly 24 hours ago and now I have a great news for you, you just gained 200 subs within 24 hrs . I totally appreciate your determination to upload quality maths questions almost everyday . 👍

[coolmangame]

I like how he just multiplies them and the system is beaten right away lol

[MathElite]

I know it's going to be a great video but no thumbnail as of now
Edit: YouTube bug it seems

[dzpismu]

The least probability by common sense turns out the best solution. Without seeing the result I would never try to multiply the three equations. Really nice!

[nrybkip]

This is the best of this polynomial system )) thanks for your hard work and video )

[bamdadtorabi]

Hey Syber! Nice work on this one. I just saw this and immediately thought about using a, b and c instead of 1, 2 and -3 in the problem. Now using your method, which you multiplied the three equations, we are left with a quadratic in the general case, but the coefficient of t^2 is (a+b+c). As with your example, this cancels to zero and generally if a+b+c=0 we get some nice solutions which are proportional to a, b and c. My question is however: Do you think we can find any a, b and c with a non zero sum, that also gives us some rational answers for xyz?

Sorry thats a bit too much😅 But it'd make a nice number theory problem! Anyways thanks for always making great math content!😁

[arshsverma]

even though I couldn't watch the premiere, it's still an exciting problem

[strikerstone]

xyz=x^3-1=y^3-2=z^3+3, so we get y={x^3+1)^(1/3), and z= (x^3-4)^(1/3), substituting this in the first eqn and we get a easy equation

[mcwulf]

Just multiply them all together. The 9-ic tems cancel as do the 6-ic. You are left with (xyz)^3 = 1/7. Substitute to find the three variables.

[laokratis]

Just for the fun of it, a short elaboration of the variation of the method. Since x+y+z=0 is valid for this system, the substitutions y=kx and z=mx imply that m=-(k+1). Then by substracting the second from the first equation and the third from the first we get x^3(1-k^3)=-1 and x^3(1+(1+k)^3)=4. Dividing them we eliminate x to find k, (1-k^3)/(1+(1+k)^3)=-1/4 which can be reduced to the form (k-2)(k^2+k+1)=0 yielding the real solution k=2 and two imaginary solutions. Then substituting k=2 in the first gives x=1/(7)^(1/3).

[Qermaq]

Folks, sometimes the thumbnail is slow to upload. On videos where the thumbnail is just a frame of the video that's not an issue, but here the thumbnail is a separate upload that has to resolve in YouTube's wonky system. It'll show up in time. :D

[ramaprasadghosh]

Well 2(x^3 + y^3 + z^3 -3xyz )
= (x+y+z)((x-y)^2 + (y-z)^2 +(z-x)^2 )
In this problem
(x^3 + y^3 + z^3 -3xyz )= 0
so either x=y=z or x+y+z= 0.
since x+y=z does not satisfy given equations,
therefore x+ y+ z= 0
.

[holyshit]

If you add these equations you will get symmetric function
With Newton's formula you can express this symmetric function in terms of elementary functions

[sahilsinghbhandari]

Again it is a nice question with a nice approach.

[PS-mhts]

x³+y³+z³-3xyz=0 does not necessarily imply x+y+z=0 (0:0:54). It can also imply x=y=z

[mathemagic]

No problem with the thumbnail. It happens with me everyday. The night I upload videos, next morning the thumbnail finds its place .. Keep it up. :)

[leecherlarry]

there is 1 real solution and 8 complex solutions:
Solve[{x^3 == x y z + 1, y^3 == x y z + 2, z^3 == x y z - 3}]

[buridivaklnagaraju]

I solved the problem by adding them and the solution I got is x=1/cuberoot of 3, y=1/cuberoot of 3, z=-2/cuberoot of 3