
#1
Apr2212, 06:09 PM

P: 7

Can any solve this x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97




#2
Apr2212, 07:17 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,388

Plan A: multiply out (x+y+z)^2 and (x+y+z)^3.
From that and the given eqations, you can get the values of yz + zx + xy and xyz. Then, you can write down a cubic equation whose roots are x y and z. Plan B: take a guess that the solution will probably be integers, and find 3 integers whose squares sum to 35. If you don't get lucky, try plan A 



#3
Apr2212, 11:19 PM

P: 145

5, 3 & 1 work.... now about there order?




#4
Apr2912, 05:14 AM

P: 7

Linear simultaneous equation
You are correctbut why don't you show your working?




#5
Apr2912, 12:25 PM

P: 145

x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97
Honestly, I just "guess and checked". 5+(3)+(1)=1 5^2+(3)^2+(1)^2=35 5^3+(3)^3+(1)^3=97 There isn't a way to find the order as far as I can see. 



#6
Apr2912, 07:25 PM

Mentor
P: 11,989





#7
Apr3012, 11:04 AM

P: 7

Am sorry that this is post in the wrong position but this is the solution x^2+y^2+z^2=(x+y+z)^22(xy+yz+zx)=35 substitute x+y+z=1 we have (1)^22(xy+yz+zx)=35 therefore (xy+yz+zx)=17
also x^3+y^3+z^3=(x+y+z)^33(x+y+z)(xy+yz+zx)+3(xyz)=97 therefore substitute we have (1)^33(1)(17)+3(xyz)=97 which give (xyz)=15 now factor of 15 whose sum is 1 give 5,1,3 


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