<s>[INST] Understanding algebraic manipulation If $a+b+c \neq 0 $ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of:
$$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$
What confusing me here, is the not so obvious hint which is given with the problem,which says that that form could be written as:
$$3- \frac{a^2}{a^2+ab+ca} - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}$$
But how is this possible?
 [/INST] André's approach is pretty simple. In case you really must follow the hint, here's how to get it: note, for example, that in the first fraction,
$$\frac{ab+ca}{a^2+ab+ca}$$
the only thing that is missing from having the numerator exactly equal to the denominator is an $a^2$; so adding it and taking it away again we get
$$\frac{ab+ca}{a^2+ba+ca} = \frac{ab+ca+a^2-a^2}{a^2+ba+ca} = \frac{a^2+ab+ca}{a^2+ab+ca} - \frac{a^2}{a^2+ba+ca} = 1-\frac{a^2}{a^2+ba+ca}.$$
The same thing can be done with
$$\frac{ab+cb}{b^2+ab+bc}$$
by adding and subtracting $b^2$ to the numerator. And also with
$$\frac{ac+cb}{c^2+ac+bc}$$
by adding and subtracting $c^2$ to the numerator. Doing all three, we have:
$$\begin{align*}
\frac{ab+ca}{a^2+ba+ca} &+ \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc} \\&= \left(1 - \frac{a^2}{a^2+ba+ca}a\right)\\
&\qquad\quad\mathop{+}\left(1 - \frac{b^2}{b^2+ab+bc}\right)\\
&\qquad\quad\mathop{+}\left(1 - \frac{c^2}{c^2+ac+bc}\right)\\
&= 3 - \frac{a^2}{a^2+ba+ca}a - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc},
\end{align*}$$
which is exactly the formula your hint gives. 
As to how the hint gives the answer any better than the straightforward method suggested by André (which would be my own approach, to simplify first), I couldn't tell you, but that is how the equation can be rewritten in that form.
</s>