ALG-B1-M02-P025 Factoring a Cyclic Fifth-Degree Expression

The expression changes sign when variables are swapped, so expect the factor \((a-b)(b-c)(c-a)\).

Let the expression be \(E\). When \(a=b\), \(b=c\), or \(c=a\), it becomes zero, so \(E\) is divisible by \((a-b)(b-c)(c-a)\).

The remaining factor has degree \(2\) and is symmetric, so it has the form \(\alpha(a^2+b^2+c^2)+\beta(ab+bc+ca)\). Put \(c=0\). Then \(E=a^4b-b^4a=ab(a-b)(a^2+ab+b^2)\), while \((a-b)(b-c)(c-a)=-(ab)(a-b)\). Hence the remaining factor is \(-(a^2+ab+b^2)\) when \(c=0\).

Thus \(\alpha=-1\), \(\beta=-1\), and

\[E=-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca).\]

This is for strong students: the factor is guessed from symmetry, and the remaining factor is found by a careful substitution.