You can consider the polynomial which has x y z as roots, and use viete's formulas from there, this is the standard method when dealing with symetric sistems of equations
Denote P1=x+y+z=3 P2=x^2+y^2+z^2=3 P3=x^3+y^3+z^3=3 Some elementary symmetric polynomials are E1=x+y+z=P1 E2=xy+yz+zx E3=xyz You can use two Newton-Girard identities P1^2 = P2 + 2E2 P3 - 3E3 = P1(P2 - E2) Substituting known values 9=3+2E2, E2=3 3-3E3=0, E3=1 -E1, E2, -E3 are the Vieta formulas for coefficients of x^2, x and 1 in a cubic with roots x, y and z x^3-3x^2+3x-1=0 (x-1)^3=0 So the roots x, y and z are all 1
I solved it a little differently. I started with "suppose x = y" and then replace y with x in all 3 equations, from there it's not hard to solve (use the first equation to solve for z, then square both sides, substitute that into the second equation, and you see that x=1, (1,1,1) is the only solution under that circumstance From there, all remaining solutions would have to have x != y != z !=x. We already found all the solutions when any of the values are equal so we know they can't be equal. WLOG we can say: x is smallest, then y, then z so x = a y = a + b z = a + b + c where a can be any real number but b and c must be positive substitute that into the first two equations and from there it's not too hard to solve. You will see that b and c must both be equal to zero. but neither of them can be equal to zero, it's a contradiction, so there are no solutions other than the one we found earlier (1,1,1)
Take only first two equations (1) x + y + z = 3 (2) x² + y² + z² = 3 From (1) (3) z = 3 - x - y Substitute (3) in (2) (4) x² + y² + (3 - x - y)² = 3 Find function minimum f(x) = x² + y² + (3 - x - y)² ∂f/∂x = 0 ∂f/∂y = 0 or (5) 2x - 2(3 - x - y) = 0 (6) 2y - 2(3 - x - y) = 0 Subtract (5) - (6): 2x - 2y = 0 => x = y and 2x - 2(3 - x - x) = 0 2x - 6 + 4x = 0 => x = 1 Therefore min f(x) = 1² + 1² + (3 - 1 - 1)² = 3 when x = 1 and y = 1 and x = 1, y = 1 is solution of equation (4) and x = 1, y = 1, z = 1 is solution of system (1) - (2). P.S. The equation x³ + y³ + z³ = 3 is redudant
Just substract 2 times equation 1 from equation 2, so you have: x²-2x + y²-2y + z²-2z = -3 Add 1+1+1 on each side: x²-2x+1 + y²-2y+1 + z²-2z+1 = 0 Factorize: (x-1)² + (y-1)² + (z-1)² = 0 Each term is greater or equal than 0, so as the sum is 0 they must all be 0, so x=y=z=1.
@Altair705 I think yours is the best solution. It gets to the answer quickly and shows that the third equation is not necessary. In fact if the third equation was equal to anything other than 3 you'd have an inconsistent system!
Good point, I didn't consider complex solutions. I reworked my method and made the following variable changes: X=x-1, Y=y-1 and Z=z-1. Then it's easy to show that the equations become: X+Y+Z=0 X²+Y²+Z²=0 X³+Y³+Z³=0 By substituting Z=-X-Y into equation 2 and developing, you get: X²+Y²+X.Y=0 As X²+Y² is -Z², you deduce Z²=X.Y. In a similar way, X²=Y.Z, and Y²=X.Z. From Z²=X.Y, you can get Z³=X.Y.Z. In a similar way, X³=X.Y.Z, and Y³=X.Y.Z So: X³=Y³=Z³=X.Y.Z By subsituting into equation 3, it comes 3X³=0. So now it's clear that X=Y=Z=0 are the only solution, so x=y=z=1. What's more interesting is to see the solutions we could get by considering only the first 2 equations. X²=Y.Z and Y²=X.Z still stand. By dividing the first equation by the second (assuming non-zero solutions): (X/Y)²=Y/X, so (X/Y)³=1. So X/Y, X/Z and Y/Z are necessarily a third root of unity (1, j or j²). So from there (by skipping a few steps), you can deduce that the solutions for (x, y, z) are under the form: (1+α, 1+j.α, 1+j².α), where α is any complex number. That was quite a fun one in the end!
i did it the rather hard way from (x + y + z)² = 9 we get xy + xz + yz = 3 then (x + y + z)³ = 27 x³ + y³ + z³ + 3(x + y)(xy + xz + yz + z²) = 27 3 + 3(x + y)(3 + z²) = 27 (x + y)(3 + z²) = 8 (x + y + z)(3 + z²) = 8 + 3z + z³ 9 + 3z² = 8 + 3z + z³ z³ - 3z² + 3z - 1 = 0 (z - 1)³ = 0 z = 1 by symmetry the solution is x = y = z = 1
Third equation is redundant , Just a smokescreen. Without any calculation sphere x^2+y^2+z^2=3 tangents plane x+y+z=3 at x=y=z=1 -visible at a glance.If you want calculations - see @Altair705 below.
You can consider the polynomial which has x y z as roots, and use viete's formulas from there, this is the standard method when dealing with symetric sistems of equations
nice
Denote
P1=x+y+z=3
P2=x^2+y^2+z^2=3
P3=x^3+y^3+z^3=3
Some elementary symmetric polynomials are
E1=x+y+z=P1
E2=xy+yz+zx
E3=xyz
You can use two Newton-Girard identities
P1^2 = P2 + 2E2
P3 - 3E3 = P1(P2 - E2)
Substituting known values
9=3+2E2, E2=3
3-3E3=0, E3=1
-E1, E2, -E3 are the Vieta formulas for coefficients of x^2, x and 1 in a cubic with roots x, y and z
x^3-3x^2+3x-1=0
(x-1)^3=0
So the roots x, y and z are all 1
nice!
Just by inspection (x,y,z)=(1,1,1)
Algebraicly we can use Newton,-Girard formula.
Absolutely x,y,z = 1 😁😁😁
I solved it a little differently.
I started with "suppose x = y" and then replace y with x in all 3 equations, from there it's not hard to solve (use the first equation to solve for z, then square both sides, substitute that into the second equation, and you see that x=1, (1,1,1) is the only solution under that circumstance
From there, all remaining solutions would have to have x != y != z !=x. We already found all the solutions when any of the values are equal so we know they can't be equal. WLOG we can say:
x is smallest, then y, then z
so x = a
y = a + b
z = a + b + c
where a can be any real number but b and c must be positive
substitute that into the first two equations and from there it's not too hard to solve. You will see that b and c must both be equal to zero. but neither of them can be equal to zero, it's a contradiction, so there are no solutions other than the one we found earlier (1,1,1)
Interesting!
😎👏😎👍👍👍
Take only first two equations
(1) x + y + z = 3
(2) x² + y² + z² = 3
From (1)
(3) z = 3 - x - y
Substitute (3) in (2)
(4) x² + y² + (3 - x - y)² = 3
Find function minimum
f(x) = x² + y² + (3 - x - y)²
∂f/∂x = 0
∂f/∂y = 0
or
(5) 2x - 2(3 - x - y) = 0
(6) 2y - 2(3 - x - y) = 0
Subtract (5) - (6):
2x - 2y = 0 => x = y
and
2x - 2(3 - x - x) = 0
2x - 6 + 4x = 0 => x = 1
Therefore
min f(x) = 1² + 1² + (3 - 1 - 1)² = 3 when x = 1 and y = 1
and
x = 1,
y = 1
is solution of equation (4) and
x = 1,
y = 1,
z = 1
is solution of system (1) - (2).
P.S. The equation
x³ + y³ + z³ = 3
is redudant
Just substract 2 times equation 1 from equation 2, so you have:
x²-2x + y²-2y + z²-2z = -3
Add 1+1+1 on each side:
x²-2x+1 + y²-2y+1 + z²-2z+1 = 0
Factorize:
(x-1)² + (y-1)² + (z-1)² = 0
Each term is greater or equal than 0, so as the sum is 0 they must all be 0, so x=y=z=1.
@Altair705 I think yours is the best solution. It gets to the answer quickly and shows that the third equation is not necessary.
In fact if the third equation was equal to anything other than 3 you'd have an inconsistent system!
Nice! For reals, what you wrote is correct but the result does not always follow for complex numbers
Good point, I didn't consider complex solutions. I reworked my method and made the following variable changes: X=x-1, Y=y-1 and Z=z-1. Then it's easy to show that the equations become:
X+Y+Z=0
X²+Y²+Z²=0
X³+Y³+Z³=0
By substituting Z=-X-Y into equation 2 and developing, you get:
X²+Y²+X.Y=0
As X²+Y² is -Z², you deduce Z²=X.Y.
In a similar way, X²=Y.Z, and Y²=X.Z.
From Z²=X.Y, you can get Z³=X.Y.Z. In a similar way, X³=X.Y.Z, and Y³=X.Y.Z
So: X³=Y³=Z³=X.Y.Z
By subsituting into equation 3, it comes 3X³=0. So now it's clear that X=Y=Z=0 are the only solution, so x=y=z=1.
What's more interesting is to see the solutions we could get by considering only the first 2 equations.
X²=Y.Z and Y²=X.Z still stand. By dividing the first equation by the second (assuming non-zero solutions):
(X/Y)²=Y/X, so (X/Y)³=1.
So X/Y, X/Z and Y/Z are necessarily a third root of unity (1, j or j²).
So from there (by skipping a few steps), you can deduce that the solutions for (x, y, z) are under the form:
(1+α, 1+j.α, 1+j².α), where α is any complex number.
That was quite a fun one in the end!
x = 1, y = 1, z = 1
(1,1,1) idfk the rest
(1, 1, 1)
Me: 1
If you proctored in 1973 then you must be at least 70!
You’re wrong 70! Years in more than the age of the universe
I did not proctor this one. It was back in 2006 I think
@@SyberMath: Ah ha!
i did it the rather hard way
from
(x + y + z)² = 9
we get
xy + xz + yz = 3
then
(x + y + z)³ = 27
x³ + y³ + z³ + 3(x + y)(xy + xz + yz + z²) = 27
3 + 3(x + y)(3 + z²) = 27
(x + y)(3 + z²) = 8
(x + y + z)(3 + z²) = 8 + 3z + z³
9 + 3z² = 8 + 3z + z³
z³ - 3z² + 3z - 1 = 0
(z - 1)³ = 0
z = 1
by symmetry the solution is x = y = z = 1
I liked ur method and appreciate it too 😊😊🎉🎉🎉. Keep going like this
Я ждал комплексных корней, а все сошлось на единицах, что и сразу было ясно на глазок.
Я знаю. Проблема была в том, чтобы просить сложные решения
Third equation is redundant , Just a smokescreen. Without any calculation sphere x^2+y^2+z^2=3 tangents plane x+y+z=3 at x=y=z=1 -visible at a glance.If you want calculations - see @Altair705 below.
Is the third eqn redundant for complex numbers, too?
@@SyberMath It is another story. You did not do it either.
x = u + 1
y = v + 1
z = w + 1
(u + 1) + (v + 1) + (w + 1) = 3
u + v + w = 0
(u + 1)² + (v + 1)² + (w + 1)² = 3
u² + v² + w² + 2(u + v + w) + 3 = 3
u² + v² + w² = 0
(u + 1)³ + (v + 1)³ + (w + 1)³ = 3
u³ + v³ + w³ + 3(u² + v² + w²)
+ 3(u + v + w) + 3 = 3
u³ + v³ + w³ = 0
B = u + v + w = 0
C = uv + uw + vw
D = uvw
t³ - 0t² + Ct - D = 0
t³ = -Ct + D
u³ = -Cu + D
v³ = -Cv + D
w³ = -Cw + D
0 = 0 + 3D => D = 0
t³ + Ct = 0
(u + v + w)² = u² + v² + z² + 2(uv + uw + vw)
C = uv + uw + vw = 0
t³ + Ct = 0
t³ = 0
(u, v, w) = (0, 0, 0)
*(x, y, z) = (1, 1, 1)*
A really nice way to approach it!
@@SyberMath Yes I think so. Thanks. I will finish it.
u² + v² + w² = 0 => u = 0, v = 0, w = 0 => x = 1, y = 1, z = 1