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### Sum of Powers of Roots of Polynomial Equations

```Date: 01/02/2007 at 10:29:58
From: Denise
Subject: Sum of powers of roots

I am an Algebra II math team coach looking for a quick trick to show
my students how to find the sum of powers of roots for third degree
polynomial equations.  For example, given a third-degree polynomial
equation with roots a, b, and c, find a^3 + b^3 + c^3 or a^4 + b^4 +
c^4.

Must they go through Tartaglia's formula to actually find the roots or
is there a simpler way using the sum and product of roots?  We know
how to find a^3 + b^3 and a^4 + b^4 when given the sum and product of
two roots.  Is there a similar approach for three roots?  I have tried
analyzing the expansion of (a + b + c)^3 but I'm not seeing a way to
find just a^3 + b^3 + c^3 knowing the sum and product of roots.  I
have not tried (a + b + c)^4, yet.

```

```
Date: 01/02/2007 at 15:17:15
From: Doctor Pete
Subject: Re: Sum of powers of roots

Hi Denise,

Let us define

S[n] = a^n + b^n + c^n,

where a, b, c are roots of the cubic

x^3 + px^2 + qx + r.

Consequently,

p = -(a+b+c)
q = ab + bc + ca
r = -abc.

Now we are interested in the relationship between S[n] and p, q, r.
We have

-S[n]p = (a^n + b^n + c^n)(a + b + c)
= S[n+1] + a^n (b+c) + b^n (c+a) + c^n (a+b).

But it is not clear how to proceed with the remaining terms.  However,
note

S[n-1]q = (a^(n-1) + b^(n-1) + c^(n-1))(ab + bc + ca)
= a^n (b+c) + b^n (c+a) + c^n (a+b)
+ a^(n-1) bc + b^(n-1) ca + c^(n-1) ab
= -S[n]p - S[n+1] + abc S[n-2].

Equivalently,

S[n+1] = -(S[n]p + S[n-1]q + S[n-2]r).

This rather fascinating (and easy to remember) identity is the key to
determining the value of the sum of the higher powers of the roots of
a cubic.  But we need to establish some initial values.  Clearly

S[0] = a^0 + b^0 + c^0 = 3,
S[1] = a + b + c = -p,

but we still need S[2].  This is easy enough, since

p^2 = (-(a + b + c))^2
= a^2 + b^2 + c^2 + 2(ab + bc + ca)
= S[2] + 2q,

so

S[2] = p^2 - 2q.

Therefore we see that

S[3] = -(S[2]p + S[1]q + S[0]r)
= -(p^3 - 2pq - pq + 3r)
= -p^3 + 3pq - 3r.

In turn, we can write

S[4] = -(S[3]p + S[2]q + S[1]r)
= -(-p^4 + 3p^2 q - 3pr + p^2 q - 2q^2 - pr)
= p^4 - 4p^2 q + 4 pr + 2q^2.

Of course, higher powers quickly become complicated to compute.

The identity we established suggests two questions:

1)  Is the identity true for negative values of n; i.e., can we
compute the sum of the reciprocals of the roots of a cubic?

2)  Can the identity be generalized for higher-degree polynomials?  In
particular, if we let

r[1], r[2], ..., r[m]

be the roots of a degree-m monic polynomial

F[x] = c[0] + c[1]x + c[2]x^2 + ... + c[m-1]x^(m-1) + x^m

and

S[n,m] = r[1]^n + r[2]^n + ... + r[m]^n,

is it true that

S[n+1,m] = -(S[n,m]c[m-1] + S[n-1,m]c[m-2] + ... + S[n-m+1,m]c[0])?

If so, how do we prove it?  If not, what is the correct relationship?

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials

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