Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  22x37 -  x

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   4x³⁷ - x  =   x • (4x³⁶ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  4x³⁶ - 1 

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 :  4  is the square of  2 
Check : 1 is the square of 1
Check :  x³⁶  is the square of  x¹⁸ 

Factorization is :       (2x¹⁸ + 1)  •  (2x¹⁸ - 1) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  2x¹⁸ + 1 

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 :  2  is not a cube !!

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

Trying to factor as a Difference of Squares :

 3.4      Factoring:  2x¹⁸ - 1 

Check :  2  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:

 3.5      Factoring:  2x¹⁸ - 1 

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 :  2  is not a cube !!

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

Final result :

  x • (2x18 + 1) • (2x18 - 1)