Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  (22•5d6) -  20

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   20d⁶ - 20  =   20 • (d⁶ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  d⁶ - 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 : 1 is the square of 1
Check :  d⁶  is the square of  d³ 

Factorization is :       (d³ + 1)  •  (d³ - 1) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  d³ + 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 :  1  is the cube of   1 
Check :  d³ is the cube of   d¹

Factorization is :
             (d + 1)  •  (d² - d + 1) 

Trying to factor by splitting the middle term

 3.4     Factoring  d² - d + 1 

The first term is,  d²  its coefficient is  1 .
The middle term is,  -d  its coefficient is  -1 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1 

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -1 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Trying to factor as a Difference of Cubes:

 3.5      Factoring:  d³-1 

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 :  1  is the cube of   1 
Check :  d³ is the cube of   d¹

Factorization is :
             (d - 1)  •  (d² + d + 1) 

Trying to factor by splitting the middle term

 3.6     Factoring  d² + d + 1 

The first term is,  d²  its coefficient is  1 .
The middle term is,  +d  its coefficient is  1 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1 

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   1 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Final result :

  20•(d+1)•(d2-d+1)•(d-1)•(d2+d+1)