Linear simultaneous equation

isaac200

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

AlephZero

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

huntoon

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

isaac200

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

huntoon

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.

Redbelly98

Why do you think order would matter here? And, why is this posted in "Mechanical Engineering"?

isaac200

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