The sum of three numbers is 6, the sum of their squares is 8, and 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

a^{2} + b^{2} + c^{2} = 8

a^{3} + b^{3} + c^{3} = 5

We will find the (monic) cubic equation whose roots are a, b, and c.

If cubic equation x^{3} − Ax^{2} + 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} − (a^{2} + b^{2} + c^{2})] = 14.

Hence a, b, c are roots of x^{3} − 6x^{2} + 14x − C = 0, and we have

a^{3} − 6a^{2} + 14a − C = 0

b^{3} − 6b^{2} + 14b − C = 0

c^{3} − 6c^{2} + 14c − C = 0

Adding, we have (a^{3} + b^{3} + c^{3}) − 6(a^{2} + b^{2} + c^{2}) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0.

Hence C = 41/3, and x^{3} − 6x^{2} + 14x − 41/3 = 0.

Multiplying the polynomial by x, we have x^{4} − 6x^{3} + 14x^{2} − 41x/3 = 0. Then

a^{4} − 6a^{3} + 14a^{2} − 41a/3 = 0

b^{4} − 6b^{3} + 14b^{2} − 41b/3 = 0

c^{4} − 6c^{3} + 14c^{2} − 41c/3 = 0

Adding, we have (a^{4} + b^{4} + c^{4}) − 6(a^{3} + b^{3} + c^{3}) + 14(a^{2} + b^{2} + c^{2}) − 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.

Note that we did not need to actually calculate a, b, and c in order to determine the sum of their fourth powers. In fact, one of the numbers is real; the other two are complex conjugates; see below. The approximate values of the numbers are 2.67770, and 1.66115 ± 1.53116*i*.

Using the above approach, we can show that if

a + b + c = r

a^{2} + b^{2} + c^{2} = s

a^{3} + b^{3} + c^{3} = t

then a, b, c are roots of x^{3} − rx^{2} + ½(r^{2} − s)x + (½r(3s − r^{2}) − t)/3 = 0.

It then follows that

a^{4} + b^{4} + c^{4} | = 4rt/3 − ½s(r^{2} − s) + r^{2}(r^{2} − 3s)/6. |

= (r^{4} − 6r^{2}s + 3s^{2} + 8rt)/6. |

Let f(n) = a^{n} + b^{n} + c^{n}, where n is a positive integer.

Then, given the equation x^{3} − 6x^{2} + 14x − 41/3 = 0, we can multiply by x^{n} 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 a^{5} + b^{5} + c^{5}, a^{6} + b^{6} + c^{6}, and so on.

- Polynomial Roots
- Sums of Powers in Terms of Symmetric Functions
- The Geometry of the Cubic Formula
- How to discover for yourself the solution of the cubic

Source: Traditional