SOLUTION: Hello Please help me dividing this polinomials using (Long division) (a^3+b^3+c^3-3abc)÷(a+b+c) I'm a bit slow in algebra so please answer my problem step-by-step. Hope you

Using only the digits 0,1,2,3, write down all 
possible three digit numbers with sum of digits 3,
allowing numbers to begin with 0, in order from 
largest to smallest:

300
210
201
120
111
102
030
021
012
003

in each, use the digits of each 3-digit number 
in the order they appear as exponents of 
a, b, and c respectively, as this scheme shows:

300 --> a3b0c0 = a³
210 --> a2b1c0 = a²b
201 --> a2b0c1 = a²c
120 --> a1b2c0 = ab²
111 --> a1b1c1 = abc
102 --> a1b0c2 = ac²
030 --> a0b3c0 = b³
021 --> a0b2c1 = b²c
012 --> a0b1c2 = bc²
003 --> a0b0c3 = c³

Rewrite the dividend a³+b³+c³-3abc putting in zero
place-holders for each of those not represented in
that dividend, in that order.  That is, the dividend 
becomes:

a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³

So we start with this:
         _______________________________________________________________ 
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³

Then we divide a³ by a, getting a², and we write that as the 1st term of the quotient.

          a²                                                            
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
         

Then we multiply that by each term of the divisor and place each product
under the term that it is like, and draw a line under it:

          a²                                                            
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
          a³ +  a²b +  a²c

Then we subtract and bring EVERY term down

          a²                                                            
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
          a³ +  a²b +  a²c
               -a²b -  a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³

Then we divide -a²b by a, getting -ab, and we write that as the 2nd term of the quotient:

          a² - ab                                                       
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
          a³ +  a²b +  a²c
               -a²b -  a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³

Then we multiply that by each term of the divisor and place each product
under the term that it is like, and draw a line under it, subtract and bring
EVERY term down:

          a² - ab                                                       
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
          a³ +  a²b +  a²c
               -a²b -  a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
               -a²b        -  ab² -  abc
                      -a²c +  ab² - 2abc + 0ac² +  b³ + 0b²c + 0bc² + c³

Keep doing that, and end up with this:

          a² - ab - ac + b² - bc  + c²                                  
a + b + c)a³ + 0a²b + 0a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
          a³ +  a²b +  a²c
               -a²b -  a²c + 0ab² - 3abc + 0ac² +  b³ + 0b²c + 0bc² + c³
               -a²b        -  ab² -  abc
                      -a²c +  ab² - 2abc + 0ac² +  b³ + 0b²c + 0bc² + c³
                      -a²c        -  abc -  ac²
                              ab² -  abc +  ac² +  b³ + 0b²c + 0bc² + c³
                              ab²               +  b³ +  b²c
                                  -  abc +  ac² - 0b³ -  b²c + 0bc² + c³
                                  -  abc              -  b²c -  bc²
                                            ac² - 0b³ + 0b²c +  bc² + c³
                                            ac²              +  bc² + c³
                                                                       0


So the answer is: a² - ab - ac + b² - bc  + c²

Edwin