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.?
For my answer, will write this equation as:


(a/(a+b)) + (b/(b+c)) + (c/(c+a))





To find the least common denominator, we multiply all the denominators together:


(a+b)(b+c)(c+a)





So now we multiply each of sections by this common denominator:





(a(a+b)(b+c)(c+a)/(a+b)) + (b(a+b)(b+c)(c+a)/(b+c)) + (c(a+b)(b+c)(c+a)/(c+a))





Now we look at the equation and see that in the first section we have (a(a+b)(b+c)(c+a)/(a+b)) which has an (a + b) on top and bottom, so we can cancel that out. We can do this on each section and find:


a(b+c)(c+a) + b(a+b)(c+a) + c(a+b)(b+c)





This can be left like that, or multiplied out:


a(ab + cc + ac + bc) + b(aa + bc + ac + ab) + c(ab + bb + ac + bc)





a^2b + ac^2 + a^2c + abc + a^2b + b^2c + abc + ab^2 + abc + b^2c + ac^2 + bc^2





Now add together any common numbers:


2a^2b + ab^2 + a^2c + 2ac^2 + 3abc + 2b^2c + bc^2





This could be written as:


a(2ab + b^2 + ac + 2c^2 + 3bc) + b(2bc + c^2)


or


a(b(2a + b + 3c) + c(a + 2c)) + b(c(2b + c))








Good luck!!