Simplifying a sum with fractions of products of binomials

[Rationalist]

Homework Statement

1/(a-b)(a-c) + 1/(c-a)(c-b) + 1/(b-a)(b-c)

Homework Equations

Is there a way to simplify this? If I start multiplying out everything to get the LCD my final answer will be huge.

The Attempt at a Solution

As I said, without somehow simplifying it at the start I can multiply out, get the LCD, and end up with a very long answer. Thanks.

 

[kenewbie]

I don't think there are any ways to simplify this. At least the numerators are 1, so that part will come automaticly. And once you have it above a common denominator you will be able to simply the heck out of it me thinks.

k

 

[Rationalist]

I got (c-a)(c-b)(b-a)(b-c)+(a-b)(a-c)(b-a)(b-c)+(a-b)(a-c)(c-a)(c-b) over LCD of all six terms combined.

I then reversed the signs for (c-a)(b-a) in the first term of the numerator and factored both out.

Then I reversed the signs for (b-a)(b-c) in the middle term of the numerator and factored out (c-b)

Finally:
(b-c)+(a-b)+(c-a)/(c-a)(b-a)(b-c)

 

[Gib Z]

Rewrite the original statement as

\frac{1}{(a-b) (a-c)} + \frac{1}{ (a-c)(b-c) } - \frac{1}{(a-b)(b-c)}.

It's easy to see that they will easily have a common denominator of (a-b)(a-c)(b-c) if you multiply the numerator and denominator of each term by the factor they are missing. Once you've done that, combine the fractions and expand the resulting numerator. Many terms should cancel out leaving you with a nice simple small answer.

 

[kenewbie]

(a-b)(b-c) = ab-ac-b²+bc
(b-a)(c-b) = bc-b²-ac+ab
(a-b)(b-c) = (b-a)(c-b)

Yikes, how could I not know this? The order of multiplication doesn't matter, but I didn't see this one at all :/

k

 

[Gib Z]

We don't even need to expand them =] As long as we know that, in general, (x-y) = -(y-x) and that a negative times a negative gives a positive, then we are set!

 

[kenewbie]

Aye, I just expanded them to convince myself it worked :)

I learned something really fundamental here, thanks.

k