Distribution or Elemination ???

Open the brackets by multiplying...combine the numerators with similar denominators then take cubes common to get the (a^3+b^3)(a+b)/(a^3+b^3) + (c^3+b^3)(c+b)/(c^3+b^3) + (a^3+c^3)(c+a)/(a^3+c^3) =2(a+b+c)=2*1=2

Since it says for all solutions, take the set (1,0,0) i.e. a=1 and b=c=0. this changes the terms of the given equation into zeroes and ones only 4th and 6th terms are 1 which results in the answer being 1+1=2

just dont limit your assumptions ..... as the question states that for " ALL" real +ve numbers ....let a,b,c be a=b=c ......then our equation reduces to just fractions of known quantities .........then get everything in terms of a and we get 3a=1 so a= 1/3..............put 1/3 wherever a and then finally u get 3(1/3) + 3(1/3) =2 ..............thats it ! :)