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.
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.😎
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..
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.
@@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.
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
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.
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
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.
It was very, very, difficult to find you. Though I subscribed your RUclips, and I am Premium, I couldn't find you. RUclips 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 RUclips about it?
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.
@@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.
@@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.
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.
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.
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
![Find (x+y+z) [Harvard-MIT] Guts contest](http://i.ytimg.com/vi/Pe0s5N_jkLA/mqdefault.jpg)
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.
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.😎
Amazing how easy or difficult this channel's problems get. Thanks for the help!
I agree,the difficulty changes a lot.
You are explained very well! Continue to teaching!
Very cool. So asking for potential complex roots, they were hoping to mislead you forcing you to investigate further.
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..
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.
@@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.
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
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.
maybe 16:37 should not be "or" but be "and" consider the meaning of vieta's formula
Beautiful explanation, really liked the solving method!
You are amazing how long it take you to develop this incredible maths skills that revolutionalize the world
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
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.
It was very, very, difficult to find you. Though I subscribed your RUclips, and I am Premium, I couldn't find you. RUclips 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 RUclips about it?
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.
I think we can apply the AM-GM inequality after we have (xy+yz+zx)=x^2 + y^2 +z ^2 =3
x+y+z=3
x^2+y^2+z^2+2xy+2xz+2yz=9
xy+xz+yz=(9-3)/2=3
x(y+z)+yz=3
x(3-x)+yz=3
x^2-3x+3=yz
(x^2+y^2+z^2)(x+y+z)=9
x^3+y^3+z^3+xy(x+y)+xz(x+z)+yz(y+z)=9
xy(3-z)+xz(3-y)+yz(3-x)=6
3(xy+xz+yz)-3xyz=6
xyz=1
yz=1/x
x(y+z)+yz=3
x(3-x)+1/x=3
x^3-3x^2+3x-1=0
(x-1)^3=0
x=1
xz=1/y
y(x+z)+1/y=3
y=1
xy=1/z
z(x+y)+1/z=3
z=1
=
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.
@@Tom_TP we did it almost exactly the same lol
@@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.
@@Tom_TP i found introducing a mercenary term to be a bit confusing and unnecessary tbh
@@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.
Wow, nice, I went with liner agebra, nm
Wonderful sir..long live
Brilliantly systematic!❤
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.
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.
Excellent
A lovely method to solve this beautiful sum.Tq Sir
Great !!
Ans is 1=x=y=z by observation
Very nice.
Solved in 0.1 seconds 😂...
Me also by observation x=y=z=1
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.
@@Grecks75 ek answer to nikala km sa km🤣🤣
O caly kuloty ye koi swal ha.x=1,y=1z=1
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
*x + y + z = 3*
(x + y + z)² = 9 = x² + y² + z² + 2(xy + xz + yz) = 3 + 2(xy + xz + yz)
*xy + xz + yz = 3*
(xy + xz + yz)(x + y + z) = 9 = 3xyz + xy(x + y) + (xz)(x + z) + yz(y + z)
9 = 3xyz + x²y + xy² + x²z + xz² + y²z + yz²
9 = 3xyz + x²y + xy² + x²z + xz² + y²z + yz²
(x² + y² + z²)(x + y + z) = 9 = x³ + y³ + z³ + x²y + x²z + y²x + y²z + z²x + z²y
x²y + x²z + y²x + y²z + z²x + z²y = 6
9 = 3xyz + x²y + xy² + x²z + xz² + y²z + yz²
3xyz = 3 => *xyz = 1*
t³ - 3t² + 3t - 1 = 0
(t³ - 1) - 3t(t - 1) = 0
(t - 1)(t² - 2t + 1) = 0
(t - 1)(t - 1)² = 0
(t - 1)³ = 0 => t - 1 = 0 => t = 1
*(x, y, z) = (1, 1, 1)*