Factorizing symmetric expressions Factorize $$\sum_{\text{cyc}}a^4(b-c)$$ I used cyclic/symmetric method to find that $(a-b), (b-c), (c-a)$ are factors of the above equation but I am not able to completely factorize this. I used long division method but I found it quite lengthy so is there any shorter method than to apply long division method after find the three factors above. Cheers!

$$\sum_{cyc}a^4(b-c)=a^4(b-c)+bc(b^3-c^3)-a(b^4-c^4)=$$
$$=(b-c)(a^4+b^3c+b^2c^2+bc^3-a(b^3+b^2c+bc^2+c^3))=$$
$$=(b-c)(a(a^3-b^3)+b^2c(b-a)+bc^2(b-a)+c^3(b-a))=$$
$$=(b-c)(a-b)(a^3+a^2b+ab^2-bc^2-b^2c-c^3)=$$
$$=(b-c)(a-b)((a-c)(a^2+ac+c^2)+b(a-c)(a+c)+(a-c)b^2)=$$
$$=(a-b)(a-c)(b-c)(a^2+b^2+c^2+ab+ac+bc).$$
Another way:
Easy to see that for $a=b$ our expression is equal to $0$, which gives a factor $a-b$.
The same words we can say about $a=c$ and about $b=c$.
Thus, we got a factor $$(a-b)(a-c)(b-c)$$ and the last factor should be something quadratic, cyclic and homogeneous, which says that it should be something symmetric.  
Now, from the the coefficient before $a^4b$ we obtain something like the following:
$$(a-b)(a-c)(b-c)(a^2+b^2+c^2+k(ab+ac+bc))$$ and from here we can get the answer.
I prefer the first way for the Schur's polynomials of a big degree.
For example, factoring $$\sum_{cyc}(a^5b-a^5c)=(a-b)(a-c)(b-c)\sum_{cyc}\left(a^3+a^2b+a^2c+\frac{1}{3}abc\right)$$ much more better to get by the first way.  
