Thank you very much. Yes I was able to take it from there and finish is and find the answer was quite an elegant solution. Your help is appreciated. Thanks.
On Wed, 16 Jan 2013 09:33:19 +0100, Wasell <Wasell@example.invalid> wrote:
>On Tue, 15 Jan 2013 23:59:53 -0600, in article ><firstname.lastname@example.org>, 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 >> mess. >> >> Your help is appreciated. >> >> s > >Let x = a-b, > y = b-c, > z = c-a. > >Then the first three terms become -(x^2 + y^2 + z^2)/(xyz). > >Note that x+y = -z, > y+z = -x, > z+x = -y. > >I'm sure you can take it from there. > >HTH >/Wasell