Solving a cubic system for real solutions

This video is about solving a cubic system of equations. The problem has a twist.
If you need to post a picture of your solution or idea:
#ChallengingMathProblems #PolynomialSystems

[davidseed]

Here is a short cut.
Transform each equation to the form x+y=cbrt(z)
y+z=cbrt(x)
Subtract
x-z =cbrt(z) -cbrt(x)
But not e that the LHS and the RHS must have opposite signs. thus x-z=0, generalising
x=y=z substituting in eqn 1
(2x)^3 =x
8x^3=x
Thus x=0 or 8x^2=1.
If so, x= +/-sqrt(1/8)
EDIT
I've been challenged on the assumption that the roots are real. I offer this alternative approach.
if f(x)= x+cbrt(x) then from above f(x)=f(z). f(x) is monotonic and therefore f-1(x) is unique. it follows that z=x

[SyberMath]

This problem is not that hard but it is an interesting one, I think, for two reasons:
1) It has a symmetry
2) The solution method leads us to a situation where we can use "Completing the Square" method to narrow down the solutions!
Any thoughts? ⬇️

[davidr]

The 3 equations are completely symmetric: exchange any two variables and you get the same equations. This tells me that x=y=z which quickly leads to 0 and +/- sqrt(2)/4

[christopherellis]

On my first substitution, x=y=1, it was obvious that this was wrong. 2³ 9³ 9³
Likewise (1+0)³ (0+1)³ (1+1)³
Not sure where ½sin45 comes from

[WolfgangKais]

What about the negative root of 8k^2-1?

[bandini]

When I put this into my scientific calculator I got NAN-NANi

[mrfork]

It is easy to see that x, y and z have the same sign. So (x+y)(y+z) >= 0 and second term of (x-z)( ...) cannot be null

[MrLidless]

You can skip the first eight minutes and say by symmetry x=y=z. This is an axiom.

[Drk]

Question: where are you from, @SyberMath? Do you have an degree in Math or do u have this project because you enjoy it?

[tarunmnair]

amazing video, but i think since the system of equations is cubic, there should be 3 solutions for x, y & z. If im right, even -sqrt(2)/4 is also a valid solution considering negative values for x, y&z...

[bollyfan]

what about x = y = z = - sqrt(2) / 4 ?

[asmocak]

Burada simetri durumu oldugundan soyle bir cozum urettim :
Genelligi bozmadan, x>y>z alalim. Dolayisiyla (y+z)^3>(z+x)^3>(x+y)^3 olur. a^3 fonksiyonu artan bir fonksiyon oldugu icin esitsizligimiz soyle olur :
y+z>z+x>x+y olur. Buradan da z>y>x cikar. Ancak bu bir celiskidir. Dolayisiyla x=y=z olur. Buradan da her bir degisken icin uc degere ulasiriz : 0, sqrt(1/8), -sqrt(1/8) olur.

[Ssilki_V_Profile]

Fust you shold use simmetry of equation and claim:
x <= y <= z.

[hamster]

Im not sure if this is correct
x+y=z⅓
x=z⅓—y
2x+2y+2z=x⅓+y⅓+z⅓
(2x)³=x
8x³—x=0
x(8x—1)=0

[PrajwalKotha]

Why we haven't took -√2/4 as another solution?

[lawrencehmurcik]

missed one solution -- missed negative sqrt(2)/4

[benheideveld]

Way too complicated algebrrhea. As the equations are symmetric in x, y and z, look for obvious solutions that have x=y=z.
Now you get three times the same equation:
(2x)³ = x
8x³ - x = 0
x=0 or x² = 1/8
x=y=z=0 or x=y=z=±¼√2
Done!

[lokeshkalamalla]

why do you always leave imaginary solutions... please try to include them