Prove that $\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$ 
Prove that $$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$

To do this I unified three terms on the left side with a common denominator and then factored the numerator (with the aid of Wolfram Alpha... as the numerator is a 5th order expression).
$$\frac{b²c²(c-b)+c²a²(a-c)+a²b²(b-a)}{abc(a-b)(b-c)(c-a)}=\dots=\frac{(ab+bc+ca)(a-b)(b-c)(c-a)}{abc(a-b)(b-c)(c-a)}$$
My question is would there be a better or easier way to lead from the left side to the right side? Actually, the original question was to 'simplify' the left side without showing the right side. I am not sure if I could do so without Wolfram Alpha. So that's why I'm asking an easier way.
What I've thought of was, for example,
$$f(x):=(x-a)(x-b)(x-c)$$
$$\text{LHS}=abc\left(\frac1{a^2f'(a)}+\frac1{b^2f'(b)}+\frac1{c^2f'(c)}\right)$$
which didn't help.

We have to prove that
$$ \sum_{cyc}\frac{bc}{a(a-b)(a-c)}=\sum_{cyc}\frac{1}{a} $$
or:
$$ \sum_{cyc}\frac{bc-(a-b)(a-c)}{a(a-b)(a-c)} = \sum_{cyc}\frac{b+c-a}{(a-b)(a-c)}=0.$$
Since
$$ 0 = \sum_{cyc}\left(\frac{1}{a-b}+\frac{1}{a-c}\right)=\sum_{cyc}\frac{2a-b-c}{(a-b)(a-c)} $$
it is enough to prove that
$$ \frac{a}{(a-b)(a-c)}+\frac{b}{(b-c)(b-a)}+\frac{c}{(c-a)(c-b)}=0$$
or, by multiplying both sides by $(a-b)(a-c)(b-c)$,
$$ a(b-c)-b(a-c)+c(a-b) = 0 $$
that is trivial.
