# Tricky math problem

by (x)
Tags: math, tricky
 P: 323 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).
 P: 45 I'm going to ask my math teacher :)
P: 100
 Quote by (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).

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.

P: 323

## Tricky math problem

 Quote by regor60 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.
 P: 45 How do they solve it ?
 P: 31 I got the answer but i cantīt get the values os x, y and z, but i got that x.y.z = Spoiler x.y.z=1
 P: 31 Ok...that was prety hard to solve...it took many hours. I hope you have an easier way to get the answer (tricky): Spoiler x= 2 cos (20) y= 2 cos (140) z= 2 cos (260) the answer to the question is 6
 P: 31 At a glance: Spoiler x = -y-z x<2 and positive y not equal z y and z are negative x.y.z = 1
 P: 31 OK, I got it...itīs really simple to get the answer to x2+y2+z2=?.... Spoiler 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 Thatīs the tricky way...

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