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
= -256, giving:
x^6+ y^6+ z^6 = 1681/3 - 256 = 913/3.
Edited on May 31, 2007, 5:51 am