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.
 A: Claim: 
$$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$
Proof: 
Note that the Lagrange interpolating polynomial through $(a,p),(b,q),(c,r)$ is given by:
$$f(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}p+\frac{(x-a)(x-c)}{(b-a)(b-c)}q+\frac{(x-a)(x-b)}{(c-a)(c-b)}r$$
We seek the value of $f(0)$ when $p=\dfrac{1}{a},q=\dfrac{1}{b},r=\dfrac{1}{c}$, i.e. when $y=f(x)$ interpolates $y=\dfrac{1}{x}$ at $x=a,b,c$.
We then notice that $1-xf(x)$ is a degree $3$ polynomial interpolant passing through $(0,1),(a,0),(b,0),(c,0)$, and is in fact the unique such polynomial.
By inspection, one must have 
$$1-xf(x)=\frac{(x-a)(x-b)(x-c)}{(-a)(-b)(-c)}=\left(1-\frac{x}{a}\right)\left(1-\frac{x}{b}\right)\left(1-\frac{x}{c}\right)$$
Expanding to leading order,
$$1-xf(x)=1-x\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\text{ higher order terms}$$
which, upon a moment's thought, gives that $f(0)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ as desired.
Note that this method gives a natural generalisation to when we have more than just $a,b,c$.
A: Hint: $P(a) = b²c²(c-b)+c²a²(a-c)+a²b²(b-a)$ is a $3^{rd}$ degree polynomial in $a$, and has $a=b$ and $a=c$ as obvious roots, so it is divisible by $(a-b)(a-c)$. By symmetry in $a,b,c$ it is divisible by $(b-c)$ as well. Divide successively by $(a-b),(b-c),(c-a)$ using polynomial long division, and you get the quotient $ab+bc+ca$.
A: I think simpler way is to make partial fraction decomposition in $a$ of each of the terms. I.e. for the first term you get
$$
-\frac{b}{(a-c) (b-c)}+\frac{c}{(a-b) (b-c)}+\frac{1}{a}
$$
for the second
$$
-\frac{c}{(a-b) (b-c)}-\frac{c}{b (b-c)}
$$
and for the last one
$$
\frac{b}{(a-c) (b-c)}+\frac{b}{c (b-c)}
$$
So most of the terms cancel and you are left with
$$
\frac{1}{a}+\frac{b}{c (b-c)}-\frac{c}{b (b-c)}
$$
after that it is simpler already
(you can make the expansion in $b$ after that to make it even simpler before joining together the fractions)
After partial fraction in $b$ you get
$$
\frac{a+c}{a c}+\frac{1}{b}
$$
A: 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.
