Tricky math problem

[ƒ(x)]

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).

[-Aladdin-]

I'm going to ask my math teacher :)

[regor60]

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.

[ƒ(x)]

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.

[-Aladdin-]

How do they solve it ?

[LAF]

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

[LAF]

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

[LAF]

At a glance:

x = -y-z
x<2 and positive
y not equal z
y and z are negative
x.y.z = 1

[LAF]

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...