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$$\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=\frac{(a-b)(b-c)(a-c)}{abc}$$

So, I'm given a solution which is 
\begin{align}\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}&=\frac{bc(b-c)+ac(c-a)+ab(a-b)}{abc}\\
&=\frac{bc(b-c)+a^2(b-c)-a(b^2-c^2)}{abc}\\
&=\frac{(b-c)(bc+a^2-ab-ac)}{abc}\\
&=\frac{(a-b)(b-c)(a-c)}{abc}\end{align}
My question is, I don't know how to go from 1st line to 2nd line. I've no idea about it. Can anyone mind to explain for it? Thanks in advance.
My another question, is there any other way to solve this? Thanks.
 A: Here's another way to demonstrate the identity.
The expression $f(a,b,c)={b-c\over a}+{c-a\over b}+{a-b\over c}$ is homogeneous and cyclically symmetric.  When expressed as $P(a,b,c)\over abc$, homogeneity implies the polynomial $P(a,b,c)$ must be a cubic.  If we let $a=b$, we see that $f(a,a,c)={a-c\over a}+{c-a\over a}+{a-a\over c}=0$, so $a-b$ must be a factor of $P(a,b,c)$.  By cyclic symmetry, $b-c$ and $c-a$ must also be factors.  Thus
$${b-c\over a}+{c-a\over b}+{a-b\over c}={k(a-b)(b-c)(c-a)\over abc}$$
for some $k$.  To solve for $k$ we can let $(a,b,c)=(1,-1,2)$:
$${-3\over1}+{1\over-1}+{2\over2}=-3\quad\text{while}\quad{2\cdot(-3)\cdot1\over-2}=3$$
so $k=-1$.  Changing the factor $c-a$ to $a-c$ absorbs the negative sign, giving the desired result.
A: Between line $1$ and line $2$ you've got $ac(c-a)+ab(a-b)$ as terms in the numerator on the right-hand-side. These terms expand out as $ac^2-a^2c+a^2b-ab^2$ (we're not breaking the rules since we're not multiplying out the whole thing). Notice the two terms with $a^2$. We can rearrange these terms to $a^2(b-c)-a(b^2-c^2)$ (as your solution has done).
We then know by the difference of squares that $b^2-c^2=(b+c)(b-c)$ so then the numerator is $bc(b-c)+a^2(b-c)-a(b+c)(b-c)$, between lines $2$ and $3$. This means every term in the numerator  has $(b-c)$ as a factor so $(b-c)$ can be factored out. We then end up with $(b-c)(bc+a^2-a(b+c))$, which simplifies to what you've got in line $3$.
So the question is how to factor $(a^2-ab-ac+bc)$. We can do this by inspection by noticing the $a^2$ term which tells us we've got $(a+?)(a+?)$. But what two numbers have a sum of $(-b-c)$ but a product of $bc$?  The answer is $-b$ and $-c$. This tells us that $(a^2-ab-ac+bc)=(a-b)(a-c)$
And voila, out falls $bc(b-c)+ac(c-a)+ab(a-b)=(b-c)(a-b)(a-c)$.
