On Tue, 15 Jan 2013 23:59:53 -0600, stony wrote: > > Hi, > > Need a little help with this. We are simplifying the following, but > the solution is pretty lengthy and messy because of the enormous > number of factors. I was thinking that may be I am missing seeing a > pattern (some series or something). Is grunt work the only way to > solve this or is there a pattern that can simplify the whole process? > > My daughter was trying to solve this, but ended up with the mess and > then I got the same mess, but I thought there may be an easy way to > simplify this that I may be missing. > > > ((b-c)/((a-b)(a-c))) + ((c-a)/((b-c)(b-a))) + ((a-b)/((c-a)(c-b))) + > (2/(b-a)) - (2/(c-a)) > > > of course, I took all the factors in the denominator and then started > multiplying the numerator with the remaining factors to end up with a
It would be nice if you had showed us your steps. You do recognize, I hope, that a-b and b-a are just the same thing with a factor of -1 pulled out? So if you're treating (a-b) and (b-a) as different factors in finding your common denominator, you're doing much more work than you need to. In fact there are only three factors, a-b, a- c, b-c, so your common denominator will be (a-b)(a-c)(b-c).
PLEASE, with that hint, solve it on your own, and only ten look at what I've done.
First step: write all the binomial factors in the same form: