SOLUTION: We are to find the value of X^4+Y^4+Z^4 when X, Y, Z are real numbers which satisfy the following three equalities: X+Y+Z=3 X^2+Y^2+Z^2=9 XYZ=-2 Firstly it follows from the

Question 882087: We are to find the value of X^4+Y^4+Z^4 when X, Y, Z are real numbers which satisfy the following three equalities:
X+Y+Z=3
X^2+Y^2+Z^2=9
XYZ=-2
Firstly it follows from the first 2 equalities that:
XY+YZ+ZX= A
Next using:
(X^2+Y^2+Z^2)^2=X^4+Y^4+Z^4+B ( (xy)^2+(yz)^2+(zx)^2 )
we have:
X^4+Y^4+Z^4= C
Find A, B, C
I tried to solve it by substituting by X in the first 3 equations but I can't seem to get a value for X or Y to solvr it please help
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
X^4+Y^4+Z^4 = 57,
X = 1+sqrt(3),
Y = 1,
Z = 1-sqrt(3)
It doesn't matter which is which
x can be any of the three values
y can be any of the three
z can be any of the three
Those are the three values

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