SOLUTION: x+y+z =1 {{{x^2+y^2+z^2}}} = 35 {{{x^3+y^3+z^3}}} = 97

SOLUTION: x+y+z =1 {{{x^2+y^2+z^2}}} = 35 {{{x^3+y^3+z^3}}} = 97

Algebra.Com

Question 951611: x+y+z =1
= 35
= 97
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
-3, -1 & 5
RELATED QUESTIONS
y/4+z/4=z/3+x/3=x/2+y/2... (answered by ikleyn)

2x+3y-z x=2 y=-3... (answered by ikleyn)

x^2+y^2+z^2=2(x+z-1) then the value of... (answered by ikleyn)

x/6+y/2+z/6=1/2 x+y+z=5 x/3+y/2+z/6=1 (answered by ikleyn)

x/6+y/2+z/6=1/2 x+y+z=5... (answered by Boreal,MathTherapy)

x^(-2/3)*y^(k-1)*z^3/(x*y^k*z^(-2/3)) (answered by Fombitz)

(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3 -------------------------------------... (answered by Fombitz)

1/x + 2/y - 4/z = 1 2/x + 3/y + 8/z = 0 -1/x + 9/y + 10/z = 5 then solve x, y and... (answered by Fombitz)

If x>0,y>0,z>0 and x+y+z=1,prove that {{{x/2-x + y/2-y +... (answered by darq)