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Tricky math problem 
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#1
May2209, 02:53 PM

P: 323

A friend of mine showed this to me a few days ago:
x+y+z=0 x^{3}+y^{3}+z^{3}=3 x^{5}+y^{5}+z^{5}=15 x^{2}+y^{2}+z^{2}=? An answer must be supported with justification (so that I know that you didn't guess and get lucky). 


#2
May2209, 11:17 PM

P: 45

I'm going to ask my math teacher :)



#3
May2609, 09:47 AM

P: 101

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. 


#4
May2809, 03:30 PM

P: 323

Tricky math problem



#5
May2909, 03:20 AM

P: 45

How do they solve it ?



#6
Jun909, 01:18 PM

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



#7
Jun1009, 11:20 AM

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 


#8
Jun1009, 11:24 AM

P: 31

At a glance:
Spoiler
x = yz
x<2 and positive y not equal z y and z are negative x.y.z = 1 


#9
Jun1009, 03:08 PM

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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