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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?

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What have you tried? Where are you stuck? Have you tried seeing what you get when you plug in numbers (say $a = b = c = 1$)? –  JavaMan Jun 11 '11 at 3:19
    
@DJC:Please read my question carefully. –  Quixotic Jun 11 '11 at 3:20
    
I have read your question carefully. So where are you stuck? What have you tried? –  JavaMan Jun 11 '11 at 3:21
    
$\quad$@DJC:Edited. –  Quixotic Jun 11 '11 at 3:25
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3 Answers 3

up vote 7 down vote accepted

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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Thanks this is a really easy approach the hint given confused me. –  Quixotic Jun 11 '11 at 3:27
    
I can understand your answer,but I don't understand the hint given in my module (my edited question),Of-course the same thing could be done after that step given $3-1=2$. –  Quixotic Jun 11 '11 at 3:32
    
Yes,I am sure I have quoted the problem correctly. –  Quixotic Jun 12 '11 at 23:55
    
@Debanjan: The reason I wondered is that the hint seemed unreasonable in view of the obvious cancellations. –  André Nicolas Jun 13 '11 at 0:06
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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.

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It is obvious that we have to use the cancellation after the hint step,thanks for the explanation and I agree this hint is not necessary and lengthy to some extent. –  Quixotic Jun 11 '11 at 3:38
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Or just add and subtract $a^2$, $b^2$, $c^2$ to the numerators of the three fractions respectively.

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