I thought of three numbers.

Their sum is 6.

The sum of their squares is 8.

The sum of their cubes is 5.

What is the sum of their fourth powers?

(In reply to

Answer by K Sengupta)

Let the numbers be x, y and z

Then, we have:

x+y+z = 6

x^2+ y^2 + z^2 = 8

x^3+y^3+z^3 = 5

Now;

x^2+y^2+z^2

=(x+y+z)^2 + 2t, where t = xy+yz+zx

or, t = (6^2 - 8)/2 = 14

Again,

x^3 + y^3 + z^3 - 3xyz

=(x+y+z)(x^2 + y^2+ z^2 - t)

or, 5 - 3xyz = 6(8-14)

Or, xyz = (5+36)/3 = 41/3

Again:

(xy+yz+zx)^2

= u + 2xyz(x+y+z), where u = x^2*y^2 + y^2*z^2 + z^2*x^2

Or, u = 14^2 - 2*(41/3)*6 = 32

Consequently,

x^4 + y^4 + z^4

= (x^2+y^2+z^2)^2 - 2u

= 64 - 2*32

= 0

*Edited on ***May 31, 2007, 5:49 am**