Nick

Following that somewhat cryptic comment, consider the simpler similar case of the two equations a+b=6, a^2+b^2=10, and find the value of a^3+b^3.

The two equations have no real solutions, but a^3+b^3 nevertheless equals -18.

(a = 3 + 2i & b = 3 - 2i -- or the other way round.)

x^2 - 6x + P = 0, for some constant P = ab.

Substitute a, then b, and add to yield:

a^2 + b^2 - 6(a + b) + 2P = 0,

from which P = (6*6 - 10)/2 = 13.

Now multiply the quadratic by x:

x^3 - 6x^2 + 13x = 0.

Again, substitute a, then b, and add to yield:

a^3 + b^3 - 6(a^2 + b^2) + 13(a + b) = 0,

from which a^3 + b^3 = 6*10 - 13*6 = -18.

For f(n) = a^n + b^n, where n is a positive integer, this approach can be used to set up a recurrence relation:

f(n+2) = 6f(n+1) - 13f(n).

A similar method can be used to solve the original problem.

Nick

We have
(i) a plus b plus c = 6
(ii) a^2 plus b^2 plus c^2 = 8
(iii) a^3 plus b^3 plus c^3 = 5

a, b and c must be the roots of the equation
(x-a)*(x-b)*(x-c) = 0, or
x^3 - x^2*(a plus b plus c) plus x*(ab plus ac plus bc) - abc

from (i) squared - (ii), ab plus ac plus bc = 14 so we have
x^3 - 6*x^2 plus 14*x - abc = 0

Substituting the roots a for x, b for x and c for x and adding the three together therefore gives

5 - 6*8 plus 14*6 = 3*abc and abc = 15

Multiplying the cubic by x gives
x^4 - 6*x^3 plus 14*x^2 - 15*x = 0

and the same substitution process gives

(a^4 plus b^4 plus c^4) - 36 plus 112 - 75 = 0
(a^4 plus b^4 plus c^4) = -1

which I trust is the answer.

The recurrence relation is now
f(n) = 6*f(n-1) - 14*f(n-2) plus 15*(n-3)

Nick

To make one arithmetic error is understandable. To make two smacks of carelessness.

More to the point, if you don't actually calculate the values of a, b and c, there isn't any simple way of checking the answer, is there? (Simple = other than by checking each step.)