This was a little tricky to show that no other solutions were possible.
Пікірлер: 42
@socil.4718Ай бұрын
Amazing how easy or difficult this channel's problems get. Thanks for the help!
@gamingplus8625Ай бұрын
I agree,the difficulty changes a lot.
@brendanward2991Ай бұрын
I was thinking that with the cubic there can only be three solutions. (1,1,1) is a solution, found by observation. But because the equations are symmetric, this must be a triple root. So there are no other solutions.
@nymalous3428Ай бұрын
I've been practicing roots of polynomial functions for a class I'm taking. We went over in depth in the last class I took, but it's been a while now, and I'm older so the information isn't sticking like it used to. Problems like this are interesting and help to keep the knowledge around for longer. Thanks for posting.
@dorukusjr2395Ай бұрын
You are explained very well! Continue to teaching!
@golddddusАй бұрын
Accelerated Girard-Newton method: t^3 -(x+y+z)t^2 +(xy+xz+yz)t -xyz=0 t^3-3t^2+3t-xyz=0 Let's replace t with x,y,z : x^3-3x^2+3x-xyz=0 (1) y^3-3y^2+3y-xyz=0 (2) z^3-3z^2+3z -xyz=0 (3) (1)+(2)+(3) x^3+y^3+z^3-3(x^2+y^2+z^2)+3(x+y+z)-3xyz=0, 3-3*3+3*3-3xyz=0, 3=3xyz xyz = 1. Historically, Girard was the first to discover the connection between symmetric polynomials and the sum of powers solutions of a polynomial equation. It doesn't always have to be Newton Prime. At least once to be Second. Never stop learning.😎
@sev7ncry9Ай бұрын
Beautiful explanation, really liked the solving method!
@vincentmudimeli4430Ай бұрын
You are amazing how long it take you to develop this incredible maths skills that revolutionalize the world
@tgg7525Ай бұрын
We have the arithmetic mean of x,y,z which is worth 1, but also the quadratic mean of x,y,z is worth 1 (by dividing by 3 and putting the second equation at the root). According to QM => AM inequality equality case (true for x,y,z € IR, not only positive ones), we must have x=y=z. So x=y=z=1.
@Ron_DeForestАй бұрын
Very cool. So asking for potential complex roots, they were hoping to mislead you forcing you to investigate further.
@johnkabila661710 күн бұрын
Brilliantly systematic!❤
@holyshit922Ай бұрын
Sytstem of equations with symmetric polynomials In fact there are special case of symmetric polynomials called power sums There are Newton-Girard formulas which allow to express power sums in terms of elementary symmetric polynomials Elementary symmetric polynomials appear in Vieta formulas To use Vieta formulas we must have elementary symmetric polynomials
@sankararaopulla5256Ай бұрын
A lovely method to solve this beautiful sum.Tq Sir
@BRUBRUETNONOАй бұрын
Thanks for your interesting problem. Here is the way I solved it. Of course, I didn't look at your solution. Tell me, if you like mine. Greetings and keep up the good work, with sharing us challenging problems. RECALL Solve the problem with all solutions reals and complex. (i) x+y+z=3 (ii) x^2+y^2+z^2=3 (iii) x^3+y^3+z^3=3 Let's square equation (i) x+y+z=3, then (i)^2 (x+y+z)^2=3^2 x^2+y^2+z^2+2(xy+yz+zx)=3^2 and from (ii), we have 3+2(xy+yz+zx)=3^2 then (xy+yz+zx)=(3^2-3)/2=3 so (xy+yz+zx)=3 Let equation (xy+yz+zx)=3 be (iv) Let's cube equation (i) x+y+z=3, then (i)^3 (x+y+z)^3=3^3 x^3+y^3+z^3+3xy(x+y)+3yz(y+z)+3zx(z+x)+6xyz=3^3 and from (iii), we have 3+3xy(x+y)+3yz(y+z)+3zx(z+x)+6xyz=3^3 moreover from (i), we have y+z=3-x and z+x=3-y and x+y=3-z so injecting those three equalities in the above one we have 3+3xy(3-z)+3yz(3-x)+3zx(3-y)+6xyz=3^3 then 3+3^2.(xy+yz+zx)-3xyz-3xyz-3xyz+6xyz=3^3 3+3^2.(xy+yz+zx)-3xyz=3^3 and from (iv) we have 3+3^2.3-3xyz=3^3 so we have 3-3xyz=0 then xyz=1 Let equation xyz=1 be (v) From following equations (i) x+y+z=3 (iv) xy+yz+zx=3 (v) xyz=1 we recognise the symetric functions of the roots (x;y;z) of a third degree equation at^3+bt^2+ct+d=0 with the following relations between the coefficients and the roots as below (those relations being called as well as the Vieta's formulas) x+y+z=-b/a xy+yz+zx=c/a xyz=-d/a so that -b/a=3 c/a=3 -d/a=1 If we choose a=1, we have b=-3, c=3 and d=1, leading to following equation t^3-3t^2+3t-1=0 that can be written as follows (t-1)^3=0 which gives t-1=0 and finally t=1 showing that the equation has got a triple root of value 1. To conclude the system has got the only solution (x;y;z)=(1;1;1) END
@vuongdoan31829 күн бұрын
I think we can apply the AM-GM inequality after we have (xy+yz+zx)=x^2 + y^2 +z ^2 =3
@giorgibliadze11513 күн бұрын
Wow, nice, I went with liner agebra, nm
@annacerbara4257Ай бұрын
My way: I find z from the first (linear) equation z=3-(x+y) I replace it in the second and third having transformed them as a function of x+y=p xy=q in fact the system is clearly symmetrical. After having found q from the second equation by means of p, it is found from the third equation, by factoring, (p-2)^3=0 therefore p=2 triple solution ... etc.
@kereric_cАй бұрын
maybe 16:37 should not be "or" but be "and" consider the meaning of vieta's formula
@peixotof3579Ай бұрын
i used newton sum and obtained x^3 - 3x^2 + 3x - 1 = 0 and it can be factored ===> (x-1)^3 = 0 and it only can be one.
@michaelz2270Ай бұрын
Notice that since S_1, S_2, and S_3 are functions of x + y + z, x^2 + y^2 + z^2, and x^3 + y^3 + z^3, if you have any one solution (x_0,y_0,z_0) to a system x + y + z = a, x^2 + y^2 + z^2 = b, x^3 + y^3 + z^3 = c, all of the solutions will be the permutations of (x_0,y_0,z_0) since doing what was done here will lead to (t - x_0)(t - y_0)(t - z_0) = 0 in the end.
@biswambarpanda4468Ай бұрын
Wonderful sir..long live
@whaddoiknow6519Ай бұрын
Yagshemash! Put x = 1 + u, y = 1 + v, z = 1 + w. You get (with two lines of work) the same system of eqns for u, v, w, with 0's on the right. In another two lines you can show that uv + vw + wu = 0 = uvw. The only solution is u=v=w=0..
@albajasadur269416 күн бұрын
I am interested in knowing the reason for this substitution. I observed that this transformation gives very useful expressions uv + vw + wu = 0 = uvw. The information is more crucial than xyz=1. Well done.
@whaddoiknow651916 күн бұрын
@@albajasadur2694 It simplifies the first equation for x+y+z, and maintains the symmetry of the system. That the other equations simplify is due to the eventual simplicity of the solution, but could not have been guessed a priori.
@Abby-hi4sf25 күн бұрын
It was very, very, difficult to find you. Though I subscribed your KZfaq, and I am Premium, I couldn't find you. KZfaq is pushing what I didn't subscribe and not what I want, and I do not know why? I found you after intensive search. Will you notify KZfaq about it?
@otakurockleeАй бұрын
Very nice.
@casar68Ай бұрын
Great !!
@user-ql2wu5uc4pАй бұрын
Ans is 1=x=y=z by observation
@nasrullahhusnan2289Ай бұрын
Note that: • the equation are cyclical • as RHS of any one equation is integer x, y and z are integers. The reasons are as follows: * if any one of x, y, and z not an integer then all equation's RHS will not be integer * if any two of x, y, and z not an integer RHS of one equation will be an integers, but those of the other ones will not. * if x, y, z are all not integer, RHS of all equations will not be integer. (x+y+z)²=x²+y²+z²+2(xy+yz+zx) 9=3+2(xy+yz+zx) xy+yz+zx=3 (x+y+z)³=x³+y³+z³ +3(x+y+z)(xy+yz+zx)-3xyz 3³=3+3³-3xyz --> xyz=1 xyz=1 implies that • x=y=z=1 • any one of x, y, and z is 1 and the other two -1 Hence (x,y,z)={(1,1,1),(1,-1,-1),(-1,1,-1), (-1,-1,1)} A simpler way to solve is to use Newton-Girard method
This is a more traditional way to solve this system of equations, but the way Mr. Newtons has shown in the video is more beautiful.
@maxvangulik1988Ай бұрын
@@Tom_TP we did it almost exactly the same lol
@Tom_TPАй бұрын
@@maxvangulik1988 That doesn't affect what I said though. Your approach was traditional and methodical, but boring to look at. What he did in the video may not be super tight logically, but it was beautiful and more easily understandable.
@maxvangulik1988Ай бұрын
@@Tom_TP i found introducing a mercenary term to be a bit confusing and unnecessary tbh
@Tom_TPАй бұрын
@@maxvangulik1988 I can understand that. Tbh I would probably do the same as you. It's the practical way while I see Mr. Newtons' way as the artful way. Each has its own pros and cons.
@tauheedsubhani183Ай бұрын
Solved in 0.1 seconds 😂...
@user-ql2wu5uc4pАй бұрын
Me also by observation x=y=z=1
@Grecks75Ай бұрын
And how did you "solve" that there are no other solutions? 😂 I mean, in theory there could be LOTS of triples solving this system, especially when considered in the complex plane. And if you change the RHS a bit, there will be! Vieta is the shortest way to go.