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?

First of all,  we must have $$a\ne b\ne c$$
Take out the terms containing $b^2$ in the numerator
$$b^2\left(-\dfrac1{(a-b)(c-a)}+\dfrac1{(b-c)(a-b)}-\dfrac1{(c-a)(b-c)}\right)$$
$$=b^2\cdot\dfrac{-(b-c)+c-a-(a-b))}{(a-b)(b-c)(c-a}$$
$$=\dfrac{2b^2(c-a)}{(a-b)(b-c)(c-a)}$$
$$\implies\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$
$$=\dfrac2{(a-b)(b-c)(c-a)}\cdot\left( b^2(c-a)+c^2(a-b)+a^2(b-c)\right)$$
Now $b^2(c-a)+c^2(a-b)+a^2(b-c)$
$=b^2(c-a)+c^2a-bc^2+a^2b-ca^2$
$=b^2(c-a)+ca(c-a)-b(c^2-a^2)$
$=(c-a)(b^2-b(c+a)+ca)$
$=\cdots$
$=-(a-b)(b-c)(c-a)$
