# 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,…