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?

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 x⁴ − 6x³ + 14x² − 41x/3 = 0.  Then

a⁴ − 6a³ + 14a² − 41a/3 = 0
b⁴ − 6b³ + 14b² − 41b/3 = 0
c⁴ − 6c³ + 14c² − 41c/3 = 0

Adding, we have (a⁴ + b⁴ + c⁴) − 6(a³ + b³ + c³) + 14(a² + b² + c²) − 41(a + b + c)/3 = 0.
Hence a⁴ + b⁴ + c⁴ = 6×5 − 14×8 + (41/3)×6 = 0.

That is, the sum of the fourth powers of the numbers is 0.

Generalization

Using the above approach, we can show that if

a + b + c = r
a² + b² + c² = s
a³ + b³ + c³ = t

then a, b, c are roots of x³ − rx² + ½(r² − s)x + (½r(3s − r²) − t)/3 = 0.

It then follows that

a⁴ + b⁴ + c⁴  = 4rt/3 − ½s(r² − s) + r²(r² − 3s)/6. = (r⁴ − 6r²s + 3s² + 8rt)/6.

Let f(n) = aⁿ + bⁿ + cⁿ, where n is a positive integer.
Then, given the equation x³ − 6x² + 14x − 41/3 = 0, we can multiply by xⁿ and sum over the three roots to yield the following recurrence relation:

f(n+3) = 6f(n+2) − 14f(n+1) + (41/3)f(n).

Successive applications of this formula allow us to calculate  a⁵ + b⁵ + c⁵,  a⁶ + b⁶ + c⁶, and so on.

Edited on October 28, 2012, 4:06 am

Posted by Danish Ahmed Khan on 2012-10-24 14:33:51