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

x⁴ + y⁴ + z⁴ = 11
x³ + y³ + z³ = 7
x² + y² + z² = 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

Posted by K Sengupta on 2007-08-25 23:49:20