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.
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.
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.😎
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.
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
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
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.
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.
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..
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
@@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.
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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.
Amazing how easy or difficult this channel's problems get. Thanks for the help!
I agree,the difficulty changes a lot.
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.
You are explained very well! Continue to teaching!
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.😎
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
Very cool. So asking for potential complex roots, they were hoping to mislead you forcing you to investigate further.
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.
Wonderful sir..long live
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
Great !!
Terimakasih🙏
Sangat jelas, walaupun saya tidak terlalu paham bahasa Inggris.
Tulisan Anda sangat sangat rapi.
I really like that about his videos. Also there's something about math on a chalkboard that is appealing to me.
Very nice.
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
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 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.
Ans is 1=x=y=z by observation
maybe 16:37 should not be "or" but be "and" consider the meaning of vieta's formula
Solved in 0.1 seconds 😂...
Me also by observation x=y=z=1
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..
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^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.
*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)*