$$\frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-a)(b-c)} + \frac{1}{c(c-a)(c-b)} $$
I tried to get everything to the same denominator, and then simplify numerators first but it is very complicated and long if I just use brute force, to multiply all the expressions given from the previous unification of denominator.
On
$$\sum_{cyc}\frac{1}{a(a-b)(a-c)}=\sum_{cyc}\frac{bc(c-b)}{abc(a-b)(b-c)(c-a)}=\frac{(a-b)(b-c)(c-a)}{abc(a-b)(b-c)(c-a)}=\frac{1}{abc}.$$ I used the following property of the Schur's polynomial: $$\sum_{cyc}(a^2c-a^2b)=(a-b)(b-c)(c-a).$$
On
Via partial fractions, you have: $$\begin{align}\frac{1}{a(a-b)(a-c) }&= \frac{1}{bc}\cdot\frac{1}{a}+\frac{1}{b(b-c)}\cdot\frac{1}{(a-b)} + \frac{1}{c(c-b)}\cdot\frac{1}{(a-c)}\\ &=\frac{1}{abc}-\frac{1}{b(b-a)(b-c)}-\frac{1}{c(c-a)(c-b)} \end{align}$$
[Idea stolen from a deleted answer from user @auscrypt that started with partial fractions but got the coefficients wrong.]
The common denominator is $$-abc(a-b)(b-c)(c-a)$$
The numerator is
$$bc(b-c)+ca(c-a)+ab(a-b)=bc(b-c)+c^2a-ca^2+a^2b-ab^2$$
$$=(b-c)(bc+a^2-a(b+c))$$
$$=(b-c)(b(c-a)-a(c-a))=?$$