Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$ 
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$

I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?

Here we use $$ab(a-b)+bc(b-c)+ca(c-a)=a^2(b-c)+b^2(c-a)+c^2(a-b)=-(a-b)(b-c)(c-a)$$
In short we cab write $$\sum ab(a-b)=\sum a^2(b-c)=-(a-b)(b-c)(c-a)~~~~(1)$$
Then $$F=\frac{a^2+b^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-a)(b-c)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$
$$\implies F=\frac{(a^2+b^2-a^2)(c-b)+(c^2+a^2-b^2)(a-c)+(a^2+b^2-c^2)(b-a)}{(a-b)(b-c)(c-a)}$$
upon simplification in the numerator six terms: $(\pm a^3\pm b^3 \pm c^3)$ will cancel each other, we will get $6+6=12$ terms in the numerator as
$$F=\frac{-\sum a^2(b-c)-\sum ab(a-b)}{(a-b)(b-c)(c-a)}= 2,$$
on using (1).
