Simplifying a sum with fractions of products of binomials

In summary, the conversation discusses how to simplify the expression 1/(a-b)(a-c) + 1/(c-a)(c-b) + 1/(b-a)(b-c). The solution is to multiply the numerator and denominator of each term by the factor it is missing and then combine the fractions. It is also mentioned that this method is based on the fact that (x-y) = -(y-x) and that a negative times a negative gives a positive.
  • #1
Rationalist
4
0

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.
 
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  • #2
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
 
  • #3
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)
 
  • #4
Rewrite the original statement as

[tex] \frac{1}{(a-b) (a-c)} + \frac{1}{ (a-c)(b-c) } - \frac{1}{(a-b)(b-c)}[/tex].

It's easy to see that they will easily have a common denominator of [tex] (a-b)(a-c)(b-c)[/tex] 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.
 
  • #5
(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
 
  • #6
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!
 
  • #7
Aye, I just expanded them to convince myself it worked :)

I learned something really fundamental here, thanks.

k
 

What is the process for simplifying a sum with fractions of products of binomials?

The process for simplifying a sum with fractions of products of binomials involves expanding each binomial, combining like terms, and then simplifying the resulting fractions.

How do I know when the sum with fractions of products of binomials is fully simplified?

A sum with fractions of products of binomials is fully simplified when there are no more common factors between the numerator and denominator, and all fractions are in their simplest form.

What are the common mistakes to avoid when simplifying a sum with fractions of products of binomials?

Common mistakes to avoid when simplifying a sum with fractions of products of binomials include not expanding the binomials correctly, not combining like terms, and not simplifying the fractions completely.

Can I use the distributive property when simplifying a sum with fractions of products of binomials?

Yes, the distributive property can be used when simplifying a sum with fractions of products of binomials. This involves multiplying the terms inside the parentheses by the coefficient outside the parentheses.

Are there any shortcuts for simplifying a sum with fractions of products of binomials?

There are no specific shortcuts for simplifying a sum with fractions of products of binomials, but it can be helpful to factor out any common terms and use the least common denominator when combining fractions.

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