Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  (0 -  ((2•3•5n2288) • n)) -  384

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -30n²²⁸⁹ - 384  =   -6 • (5n²²⁸⁹ + 64) 

Trying to factor as a Sum of Cubes :

 3.2      Factoring:  5n²²⁸⁹ + 64 

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 :  5  is not a cube !!

Ruling : Binomial can not be factored as the difference of two perfect cubes

Final result :

  -6 • (5n2289 + 64)