Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  (2 • (x5)) +  ((24•33x2) • y3)
 Step  2  :

Equation at the end of step  2  :

  2x5 +  (24•33x2y3)

Step  3  :

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   2x⁵ + 432x²y³  =   2x² • (x³ + 216y³) 

Trying to factor as a Sum of Cubes :

 4.2      Factoring:  x³ + 216y³ 

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 :  216  is the cube of   6 
Check :  x³ is the cube of   x¹

Check :  y³ is the cube of   y¹

Factorization is :
             (x + 6y)  •  (x² - 6xy + 36y²) 

Trying to factor a multi variable polynomial :

 4.3    Factoring    x² - 6xy + 36y² 

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

 Factorization fails

Final result :

  2x2 • (x + 6y) • (x2 - 6xy + 36y2)