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

An Alternative Methodology by K Sengupta)

x^6 + y^6 + z^6 - 3*(xyz)^2

= (x^2 + y^2 + z^2)[(x^4+y^4+z^4) - u], where :

u = x^2*y^2 + y^2*z^2 + z^2*x^2 = 32, as was shown in terms

of the previous post

Thus, x^6 + y^6 + z^6 - 3*(xyz)^2

= 8*(-32)

= -256, giving:

x^6+ y^6+ z^6 = 1681/3 - 256 = 913/3.

*Edited on ***May 31, 2007, 5:51 am**