Factorizing $\sum_{abc} a(b^3-c^3)$ 
Factor
$$\sum_{abc} a(b^3-c^3)$$

The solution given is as follows:

If we consider it as $f(a)$ then when $a=b$, $f(b)= 0$; thus, $b$ is a factor of $f(a)$. And since the equation is cyclic, other two factors are also cyclic. So the factors are:
$$(a-b)(b-c)(c-a)$$

Now, my first doubt is why are we ignoring the other factors. For example, for $f(a)$, $a=b$ and $a=c$, $f(b)=f(c)=0$; thus, for $f(a)$ the factors should be $(a-b)(a-c)$.
Thus the overall factor of our polynomial then comes out to be:
$$-(a-b)^2(b-c)^2(c-a)^2$$

Why is this not considered?

Besides this doubt, the book further says, since our polynomial is in a degree $4$ and the factors are of degree $3$, we need an extra first degree cyclic expression which will be another factor. Which is $m(a+b+c)$, with $m$ being a constant.
The conclusion is that the factors of the given polynomial are
$$(a-b)(b-c)(c-a)m(a+b+c)$$

How did we guessed this extra term just by knowing that our degree was off?

 A: A different point of view: looking for a determinantal common expression, a method that often works.
If one knows Vandermonde determinants like this one
$$\det \begin{pmatrix}
1 &1 &1 \\
a &b &c \\
a^2 &b^2 &c^2\end{pmatrix}=(a-b)(b-c)(c-a) \\$$
the similitude is evident ; you just have to modify the last line:
$$\det \begin{pmatrix}
1 &1 &1 \\
a &b &c \\
a^3 &b^3 &c^3\end{pmatrix} \\$$
And now, it remains to expand this determinant in two different ways:

*

*with respect to the second line for the initial expression


*by using $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, giving:
$$\det \begin{pmatrix}
1 &0 &0 \\
a &b-a &c-a \\
a^3 &b^3-a^3 &c^3-a^3\end{pmatrix} =\det \begin{pmatrix}
1 &0 &0 \\
a &(b-a) &(c-a) \\
a^3 &(b-a)(b^2+ba+a^2) &(c-a)(c^2+ca+a^2)\end{pmatrix} $$
$$=(b-a)(c-a)(c^2+ca+a^2-b^2-ba-a^2) =(b-a)(c-a)(c-b)(c+b+a)$$
as desired.
One can find this expansion as well here.
A: (1) The thing is, the powers can be any number i.e. we can generalise that as $(a-b)^m(b-c)^m(c-a)^m$ is a factor of our expression. But our expression is of degree-four and the above expression is of degree $3m$. So, the only possible thing is $m=1$.
(2) We just knew from the offset of degree, that there is a linear factor missing, which is of course, of the form $k_1a+k_2b+k_3c$. But, the term should be cyclic. So, $k_1=k_2=k_3$, which the book considers $m$. So, the factor is $ma+mb+mc=m(a+b+c)$.
Now, that our factorisation is complete we have
$$\sum_{\text{cyc}}a(b^3-c^3)=m(a+b+c)(a-b)(b-c)(c-a)$$
You can see that the LHS has all coefficients $1$, so we must have $m=1$. This implies
$$\boxed{\sum_{\text{cyc}}a(b^3-c^3)=(a+b+c)(a-b)(b-c)(c-a)}$$
Hope this helps. Ask anything if not clear.
