Question

x+y+z=1, x(x)+y(y)+z(z)=35. x(x)(x)+y(y)(y)+z(z)(z)=96 sor.myphonehasno'raise to power'

Answer 1

recap:
x+y+z = 1
x^2 + y^2 + z^2 = 35
x^3 + y^3 + z^3 = 96

let's just take
(x+y+z)^2

(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz
1 = 35 + 2(xy+xz+yz)
xy + xz + yz = -34/2 = -17 ***

let's cube x+y+z=1
(x+y+z)^3 = x^3+y^3+z^3 + 3x^2(y+z) + 3y^2(x+z) + 3z^2(x+y)
1 = 96 + 3x^2(y+z) + 3y^2(x+z) + 3z^2(x+y)
x^2(y+z) + y^2(x+z) + z^2(x+y) = -95/3 ****

arggghhhh!

Wolfram says there are no real solutions.

http://www.wolframalpha.com/input/?i=solve+x%2By%2Bz%3D1,+x%5E2%2By%5E2%2Bz%5E2+%3D+35,+x%5E3%2By%5E3%2Bz%5E3%3D96