<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] All you need to do is "simplify" the fractions, for example by dividing "top" and "bottom" of the first one by $a$, of the second by $b$, of the third by $c$. We get
$$\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}$$
We have a common denominator $a+b+c$. The numerators add up to $2(a+b+c)$. Cancel. We get $2$.
Comment: If you find this not obvious, let's look at the first term, that is, at
$$\frac{ab+ca}{a^2+ab+ca}$$
The "top" is $a(b+c)$. The "bottom" is $a(a+b+c)$. So the fraction is
$$\frac{a(b+c)}{a(a+b+c)}$$
Divide top and bottom by $a$, or equivalently, "cancel" the $a$'s.  We are using the "algebra" version of the familiar fact that
$$\frac{2\cdot 3}{2 \cdot 5}=\frac{3}{5}$$
Additional comment The hint is strange.  True, we can rewrite the expression as 
$$\left(1-\frac{a^2}{a^2+ab+ca}\right) +\left(1-\frac{b^2}{b^2+ab+bc}\right)+\left(1-\frac{c^2}{c^2+bc+ca}\right)$$
which is equivalent to what was given. And then we could divide top and bottom by  $a$ in the first fraction, by $b$ in the second, by $c$ in the third, and end up with
$$3-\frac{a+b+c}{a+b+c}$$
But it sure seems like a lot of work when we can cancel immediately!  You are sure that you quoted the problem correctly?
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