Solution - Factoring multivariable polynomials

Step  1  :

Trying to factor as a Difference of Squares :

 1.1      Factoring:  p²⁶¹²⁴-b²¹ 

Theory : A difference of two perfect squares,  A² - B²  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A² - AB + BA - B² =
         A² - AB + AB - B² =
         A² - B²

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  p²⁶¹²⁴  is the square of  p¹³⁰⁶² 

Check :  b²¹   is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares

Trying to factor as a Difference of Cubes:

 1.2      Factoring:  p²⁶¹²⁴-b²¹ 

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 :  p²⁶¹²⁴ is the cube of   p⁸⁷⁰⁸

Check :  b²¹ is the cube of   b⁷

Factorization is :
             (p⁸⁷⁰⁸ - b⁷)  •  (p¹⁷⁴¹⁶ + p⁸⁷⁰⁸b⁷ + b¹⁴) 

Trying to factor as a Difference of Squares :

 1.3      Factoring:  p⁸⁷⁰⁸ - b⁷ 

Check :  p⁸⁷⁰⁸  is the square of  p⁴³⁵⁴ 

Check :  b⁷   is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares

Trying to factor a multi variable polynomial :

 1.4    Factoring    p¹⁷⁴¹⁶ + p⁸⁷⁰⁸b⁷ + b¹⁴ 

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

 Factorization fails

Final result :

  (p8708 - b7) • (p8708b7 + b14 + p17416)