Algebra —factorization using symmetric properties

Here is a non-trivial example of factorization using symmetric polynomials.

Question: Factorize into five factors the following expression:

a^{5}(b-c)+b^{5}(c-a)+c^{5}(a-b)+abc(b-c)(c-a)(a-b).

Solution: Observe that the given expression is symmetric, homogeneous in degree 6.

Let E=a^{5}(b-c)+b^{5}(c-a)+c^{5}(a-b)+abc(b-c)(c-a)(a-b)

If we put a=b, b=c, c=a one by one in the above expression E, the expression E collapses to zero. Hence, by the factor theorem, the following third degree, homogeneous, symmetric expression is a factor of E:

(a-b)(b-c)(c-a).

The above is just third degree so let us assume that:

E=(a-b)(b-c)(c-a)(A \sum a^{3}+B. \sum a^{2}b+C.abc).

Step 1. Put a=0, b=1, c=2 so that we get 3A+2B=-5

Step 2. Put a=2, b=3, c=0 so that we get 7A+6B=-13

Hence, solving Step 1 and Step 2, we get A=-1, B=-1.

Step 3. Put a=4,b=3,c=2 so that we get C=0.

Hence, a^{5}(b-c)+b^{5}(c-a)+c^{5}(a-b)+abc(b-c)(c-a)(a-b) is factorized into five factors as follows:

(a-b)(b-c)(c-a)(-\sum a^{3} -\sum a^{2}b) which can be further factorized as:

(a-b)(b-c)(c-a)(-a-b-c)(a^{2}+b^{2}+c^{2}).

More later,…

Please send your suggestions, comments, etc.

Nalin Pithwa

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