factorizing polynomial algebraic equations

Question

What is the method to factorize

$$a^2(b-c) + b^2(c-a) + c^2(a-b)$$

I have already tried expanding the terms and always get an extra $2c^2b$ term or another unwanted term. I've also tried to make $ (a-b-c)$ a common factor and I still end with the same problem.Is their a specific methodology to factorize such expressions and if so, please let me know and also please state if there are other similar problems and solutions.

Answer 1

see about this.. jpeg seems fairly clear. For a ternary cubic, construct the Hessian matrix of second partial derivatives. The entries are linear. Therefore the determinant is once again a ternary cubic. The simple theorem is that a ternary cubic factors completely over the complex numbers if and only if its Hessian determinant is a constant multiple of the original form. For this particular problem the Hessian determinant is zero, so the theorem applies.

It gets more difficult if the ternary cubic factors (over $\mathbb C$) as a linear times a quadratic, see Corollary 3 in the jpeg.

Answer 2

Hint: It is much easier to solve if you can observe some of the things.
Observe that $(a-b),(b-c),(c-a)$ are factors as when keeping $a=b$ the expression becomes $0$(by factor theorem) and so on.

Answer 3

$a^2(b-c)+b^2(c-a)+c^2(a-b)$.
put in maxima and then use factor( ) function you get $(a-b)(c-a)(c-b)$ as solution, see that $(a-b-c)$ wasn't there because it's not a common factor in the expression

Answer 4

The technique I learnt in high school was to break the symmetry between variables. Here, observe that $$(a-b)+(b-c)+(c-a)=0,$$ so, say $\;a-c=(a-b)+(b-c)$. We then can rewrite the expression as \begin{align} &a^2(b-c)+b^2(c-a)+c^2(a-b)=a^2(b-c)-b^2\bigl((a-b)+(b-c)\bigr)+c^2(a-b)\\ {}={}(&a^2-b^2)(b-c)+(c^2-b^2)(a-b)=(a-b)(a+b)(b-c)+(c-b)(c+b)(a-b)\\[1ex] {}={}(&a-b)(b-c)\bigl((a+b)-(c+b)= \color{red}{-(a-b)(b-c)(c-a)}. \end{align}