What is x + y + z if the following are true:

x

^{4} + y

^{4} + z

^{4} = 11

x

^{3} + y

^{3} + z

^{3} = 7

x

^{2} + y

^{2} + z

^{2} = 3

Just think of this problem as the reverse of

this one.

Since it is not mentioned that x+y+z must be real, we will work upon the assumption that the said sum may assume complex values.

Let us substitute :

A = x+y+z,

B = xy+yz+zx;

C = xyz,

D = x^2*y^2+y^2*z^2 + z^2*x^2

Then, 2B = A^2 – 3, giving:

B = (A^2 - 3)/2

Also, x^3 + y^3 + z^3 – 3C = A(x^2 + y^2 + z^2 – B)

Or, 7 - 3C = A(3-B) = A(9- A^2)/2, upon simplification.

Or, C = (A^3 - 9A + 14)/6

But,

D = B^2 - 2CA

= (A^2-3)^2/4 - 2A(A^3 - 9A + 14)/6

= (27 - 56A + 18*A^2 - A^4)/12 ............(i)

Again:

2D = (x^2 + y^2 +z^2)^2 -(x^4 + y^4 + z^4)

= 9 - 11 = -2

Or, D = -1 ..........(ii)

Equating (i) and (ii), we have:

A^4 - 18*A^2 + 56A - 39 = 0

Or, (A-1)(A^3 + A^2 - 17A + 39) = 0

Hence, either A = 1, or:

A^3 + A^2 - 17A + 39 = 0,

Solving the above cubic equation yields:

[EDIT]

A = -5.44139, 2.220693 + i* 1.49526, 2.220693 - i* 1.49526

Thus x+y+z can assume any of the four possible values and, these are :

1, -5.44139, 2.220693 + i* 1.49526, 2.220693 - i* 1.49526

__Note__: At the time of original post, there was a fundamental lacuna in solution of the cubic equation, and succinct amendments are incorporated accordingly.

[/EDIT]

*Edited on ***August 31, 2007, 9:25 am**