Question 697108
Given : (a^3+b^3+c^3-3abc)/(a+b+c)

(a^3+b^3+c^3-3abc) can be written as (a+b+c)(a^2+b^2+c^2-(ab+bc+ac))

Therefore : (a^3+b^3+c^3-3abc)/(a+b+c) = (a^2+b^2+c^2-(ab+bc+ac))

Method 2 :  a+b+c )a^3+b^3+c^3-3abc(a^2+b^2+c^2-ab-bc-ac
                   a^3+a^2b+a^2c              subtract
                 --------------------------
                   b^3+c^3-a^2b-a^2c-3abc
                   b^3+ab^2+cb^2               subtract
                 --------------------------
                   c^3-a^2b-ab^2-a^2c-cb^2-3abc
                   c^3+c^2b+c^2a                subtract
                 ---------------------------
                  -a^2b-ab^2-a^2c-cb^2-3abc-c^2b-c^2a
                  -a^2b-ab^2-abc                subtract
                 ---------------------------
                  -a^2c-cb^2-2abc-c^2b-c^2a 
                  -abc -b^2c-bc^2                subtract
                 --------------------------- 
                  -a^2c-abc+ca^2        
                 - a^2c-abc-ac^2                  subtract
                 ---------------------------
                    0


Quotient = a^2+b^2+c^2-ab-bc-ac