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Solution from http://www.qbyte.org/puzzles/p079s.html
Let the numbers be a, b, and c. Then we have
a+b+c=6
a²+b²+c²=8
a³+b³+c³=5
We will find the monic cubic equation whose roots are a, b, and c. If cubic equation x³-Ax²+Bx-C=0 has roots a, b, c, then, expanding (x-a)(x-b)(x-c), we find
A = a + b + c
B = ab + bc + ca
C = abc
Then B=ab+bc+ca=½[(a+b+c)²-(a²+b²+c²)]=14. Hence a, b, c are roots of x³-6x²+14x-C=0, and we have
a³-6a²+14a-C=0
b³-6b²+14b-C=0
c³-6c²+14c-C=0
Adding, we have (a³+b³+c³) -6(a²+b²+c²) +14(a+b+c) -3C =5-6×8+14×6-3C =0. Hence C=41/3, and x³ -6x² +14x-41/3 =0.
Multiplying the polynomial by x, we have x^4-6x³ +14x² -41x/3 =0. Then
a^4-6a³+14a²-41a/3=0
b^4-6b³+14b²-41b/3=0
c^4-6c³+14c²-41c/3=0
Adding, we have (a^4+b^4+c^4) -6(a³+b³+c³) +14(a²+b²+c²) -41(a+b+c)/3 =0.
Hence a^4+b^4+c^4 =6×5-14×8+(41/3)×6=0.
That is, the sum of the fourth powers of the numbers is 0.
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