Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  ((x6) -  (22•31x3)) -  125

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  x⁶-124x³-125 

The first term is,  x⁶  its coefficient is  1 .
The middle term is,  -124x³  its coefficient is  -124 .
The last term, "the constant", is  -125 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -125 = -125 

Step-2 : Find two factors of  -125  whose sum equals the coefficient of the middle term, which is   -124 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -125  and  1 
                     x⁶ - 125x³ + 1x³ - 125

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x³ • (x³-125)
              Add up the last 2 terms, pulling out common factors :
                     1 • (x³-125)
Step-5 : Add up the four terms of step 4 :
                    (x³+1)  •  (x³-125)
             Which is the desired factorization

Trying to factor as a Sum of Cubes :

 2.2      Factoring:  x³+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 :  x³ is the cube of   x¹

Factorization is :
             (x + 1)  •  (x² - x + 1) 

Trying to factor by splitting the middle term

 2.3     Factoring  x² - x + 1 

The first term is,  x²  its coefficient is  1 .
The middle term is,  -x  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:

 2.4      Factoring:  x³-125 

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 :  125  is the cube of   5 
Check :  x³ is the cube of   x¹

Factorization is :
             (x - 5)  •  (x² + 5x + 25) 

Trying to factor by splitting the middle term

 2.5     Factoring  x² + 5x + 25 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +5x  its coefficient is  5 .
The last term, "the constant", is  +25 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 25 = 25 

Step-2 : Find two factors of  25  whose sum equals the coefficient of the middle term, which is   5 .

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

Final result :

  (x+1)•(x2-x+1)•(x-5)•(x2+5x+25)