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?

32! I worked this out kinda backwards, but this solution will be worked forward though.

a+b+c=6. Squaring it yields a^2+b^2+c^2+2ab+2ac+2bc=36. a^2+b^2+c^2=8, so ab+ac+bc=14.

Cubing the linear sum yields a^3+b^3+c^3+3a^2*b+3a^2*c+3a*b^2+3b^2*c+3a*c^2+3b*c^2+6abc=216. Subtracting the cubes sum gets a^2*b+a^2*c+a*b^2+b^2*c+a*c^2+b*c^2+2abc=211/3. That becomes

a(ab+ac+bc)+b(ab+bc+ac)+c(ac+bc+ab)-abc=211/3
(a+b+c)(ab+ac+bc)-abc=211/3
6*14-abc=211/3 --> abc=41/3

Squaring ab+ac+bc=14 yields a^2*b^2+a^2*c^2+b^2*c^2+2a^2*bc+2a*b^2*c+2ab*c^2=196. a^2*b^2+a^2*c^2+b^2*c^2+2abc(a+b+c)=196. a^2*b^2+a^2*c^2+b^2*c^2+2(41/3)(6)=196. a^2*b^2+a^2*c^2+b^2*c^2=32

Squaring the square sums gets you,

a^4+b^4+c^4+2a^2*b^2+2a^2*c^2+2b^2*c^2=64.  a^4+b^4+c^4+32=64 --> a^4+b^4+c^4=32

Posted by np_rt on 2004-07-01 13:35:33