What is the Value of x² + y² + z² Given These Equations?

  • Context: Undergrad 
  • Thread starter Thread starter ƒ(x)
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around finding the value of x² + y² + z² given a set of equations involving x, y, and z. The context includes problem-solving techniques and potential shortcuts, with participants sharing their approaches and experiences related to the equations.

Discussion Character

  • Homework-related
  • Debate/contested
  • Exploratory

Main Points Raised

  • Some participants present the equations x+y+z=0, x³+y³+z³=3, and x⁵+y⁵+z⁵=15 as a basis for finding x²+y²+z².
  • One participant suggests consulting a math teacher for assistance.
  • Another participant expresses skepticism about the problem's complexity, indicating it could be solved using tools like Wolfram Alpha.
  • A participant mentions they have found an answer but struggles to determine the individual values of x, y, and z.
  • Some participants discuss the difficulty of the problem, with one noting it took many hours to solve.
  • There are hints of a "tricky" method to arrive at the answer, although specifics are not provided.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the problem, with varying opinions on its difficulty and the existence of shortcuts.

Contextual Notes

Some participants express uncertainty about the methods used to solve the equations and the implications of their findings, particularly regarding the values of x, y, and z.

ƒ(x)
Messages
327
Reaction score
0
A friend of mine showed this to me a few days ago:

x+y+z=0
x3+y3+z3=3
x5+y5+z5=15

x2+y2+z2=?

An answer must be supported with justification (so that I know that you didn't guess and get lucky).
 
Mathematics news on Phys.org
I'm going to ask my math teacher :)
 
ƒ(x) said:
A friend of mine showed this to me a few days ago:

x+y+z=0
x3+y3+z3=3
x5+y5+z5=15

x2+y2+z2=?

An answer must be supported with justification (so that I know that you didn't guess and get lucky).


Unless you can demonstrate that there is a clever shortcut to this problem which would raise it to the level of a brainteaser, this is just a set of simultaneous equations which is easily solvable with, for example, the Wolfram Alpha website.
 
regor60 said:
Unless you can demonstrate that there is a clever shortcut to this problem which would raise it to the level of a brainteaser, this is just a set of simultaneous equations which is easily solvable with, for example, the Wolfram Alpha website.

I had no idea that such a site even existed.
 
How do they solve it ?
 
I got the answer but i cant´t get the values os x, y and z, but i got that x.y.z =
x.y.z=1
 
Ok...that was prety hard to solve...it took many hours. I hope you have an easier way to get the answer (tricky):
x= 2 cos (20)
y= 2 cos (140)
z= 2 cos (260)
the answer to the question is 6
 
At a glance:

x = -y-z
x<2 and positive
y not equal z
y and z are negative
x.y.z = 1
 
OK, I got it...it´s really simple to get the answer to x2+y2+z2=?...
the answer is 6. and you just have to set values to x (positive) y(negative) and z(negative) and calcutale the proportion like: set x=3 y=-2 z=-1, then
^2 = 14 --> ?
^3 = 18 --> 3
^5 = 210 --> 15

so 14 . 18 . 15 = ? . 3 . 210
? = 6

That´s the tricky way...
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
5K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
20K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 15 ·
Replies
15
Views
4K