Solution - Factoring binomials using the difference of squares

Step  1  :

            m3 + n3
 Simplify   ———————
            m3 - n3

Trying to factor as a Sum of Cubes :

 1.1      Factoring:  m³ + n³ 

Theory : A sum of two perfect cubes,  a³ + b³ can be factored into  :
             (a+b) • (a²-ab+b²)
Proof  : (a+b) • (a²-ab+b²) =
    a³-a²b+ab²+ba²-b²a+b³ =
    a³+(a²b-ba²)+(ab²-b²a)+b³=
    a³+0+0+b³=
    a³+b³

Check :  m³ is the cube of   m¹

Check :  n³ is the cube of   n¹

Factorization is :
             (m + n)  •  (m² - mn + n²) 

Trying to factor a multi variable polynomial :

 1.2    Factoring    m² - mn + n² 

Try to factor this multi-variable trinomial using trial and error 

 Factorization fails

Trying to factor as a Difference of Cubes:

 1.3      Factoring:  m³ - n³ 

Theory : A difference of two perfect cubes,  a³ - b³ can be factored into
              (a-b) • (a² +ab +b²)

Proof :  (a-b)•(a²+ab+b²) =
            a³+a²b+ab²-ba²-b²a-b³ =
            a³+(a²b-ba²)+(ab²-b²a)-b³ =
            a³+0+0-b³ =
            a³-b³

Check :  m³ is the cube of   m¹

Check :  n³ is the cube of   n¹

Factorization is :
             (m - n)  •  (m² + mn + n²) 

Trying to factor a multi variable polynomial :

 1.4    Factoring    m² + mn + n² 

Try to factor this multi-variable trinomial using trial and error 

 Factorization fails

Final result :

  (m + n) • (m2 - mn + n2)
  ————————————————————————
  (m - n) • (m2 + mn + n2)