Please help.....L.C.M problem.?

How to find the L.C.M of a + b + c


___ ___ ___ ?


a+b b+c c+a





It is,a upon a+b,plus b upon b+c,plus c upon c+a





Please explain in detail with proper steps and instructions.

Please help.....L.C.M problem.?
I'm not entirely sure what you're asking for. L.C.M. normally stands for least common multiple. Sums generally don't enter into it. You'd find the L.C.M. of a set of numbers by finding the prime factors of all of them, canceling any duplicates between numbers, and then multiplying together all the ones that are left. If there are multiple repeats of any of them within a single number, they each count as a separate factor, and you can only cancel one factor in each other number for one factor in the first number.





Are you asking to find the LCM of the denominators of the three fractions in order to add them together? If so, the easiest thing is just to muliply the three denominators together, and worry about canceling things later. You have to multiply each numerator by the other two denominators, so you're just multiplying each fraction by 1 each time.





If I'm understanding your details correction, you have:


[a/(a+b)] + [b/(b+c)] + [c/(c+a)]





First you'd multiply a/(a+b) by (b+c)/(b+c) and (c+a)/(c+a). That gives you [a(b+c)(c+a)]/[(a+b)(b+c)(c+a)]





Next multiply b/(b+c) by (a+b)/(a+b) and (c+a)/(c+a).


You'll get [b(a+b)(c+a)]/[(a+b)(b+c)(c+a)]





Then multiply c/(c+a) by (a+b)/(a+b) and (b+c)/(b+c)


You get [c(a+b)(b+c)]/[(a+b)(b+c)(c+a)]





Then add the numerators together, just like adding any other fractions with the same denominator.


That gives you


[a(b+c)(c+a)] + [b(a+b)(c+a)] + [c(a+b)(b+c)]


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[(a+b)(b+c)(c+a)]





Then you can multiply out all the terms, and look for things you can cancel between the numerator and denominator.

dracaena