Answer:
0
Step-by-step explanation:
[tex]\frac{a}{(a-b)(a-c)} +\frac{b}{(b-c)(b-a)}+\frac{c}{(c-a)(c-b)}\\= \frac{a}{(a-b)(a-c)} -\frac{b}{(b-c)(a-b)}+\frac{c}{(a-c)(b-c)}\\\\= \frac{a(b-c)-b(a-c)+c(a-b)}{(a-b)(a-c)(b-c)}}\\\\= \frac{ab-ac-ab+bc+ac-bc}{(a-b)(a-c)(b-c)}}\\\\= \frac{0}{(a-b)(a-c)(b-c)}}\\\\= 0[/tex]