Solution - Factoring binomials using the difference of squares

Step  1  :

Trying to factor as a Difference of Squares :

 1.1      Factoring:  l²¹⁴²-1 

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 :  l²¹⁴²  is the square of  l¹⁰⁷¹ 

Factorization is :       (l¹⁰⁷¹ + 1)  •  (l¹⁰⁷¹ - 1) 

Trying to factor as a Sum of Cubes :

 1.2      Factoring:  l¹⁰⁷¹ + 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 :  l¹⁰⁷¹ is the cube of   l³⁵⁷

Factorization is :
             (l³⁵⁷ + 1)  •  (l⁷¹⁴ - l³⁵⁷ + 1) 

Trying to factor as a Sum of Cubes :

 1.3      Factoring:  l³⁵⁷ + 1 

Check :  1  is the cube of   1 
Check :  l³⁵⁷ is the cube of   l¹¹⁹

Factorization is :
             (l¹¹⁹ + 1)  •  (l²³⁸ - l¹¹⁹ + 1) 

Trying to factor by splitting the middle term

 1.4     Factoring  -l¹¹⁹ + 1 + l²³⁸ 

The first term is,  -l¹¹⁹  its coefficient is  1 .
The middle term is,  +1  its coefficient is  -1 .
The last term, "the constant", is  +l²³⁸ 

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 by splitting the middle term

 1.5     Factoring  1-l³⁵⁷+l⁷¹⁴ 

The first term is,  1  its coefficient is  1 .
The middle term is,  -l³⁵⁷  its coefficient is  -1 .
The last term, "the constant", is  +l⁷¹⁴ 

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:

 1.6      Factoring:  l¹⁰⁷¹-1 

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 :  l¹⁰⁷¹ is the cube of   l³⁵⁷

Factorization is :
             (l³⁵⁷ - 1)  •  (l⁷¹⁴ + l³⁵⁷ + 1) 

Trying to factor as a Difference of Cubes:

 1.7      Factoring:  l³⁵⁷ - 1 

Check :  1  is the cube of   1 
Check :  l³⁵⁷ is the cube of   l¹¹⁹

Factorization is :
             (l¹¹⁹ - 1)  •  (l²³⁸ + l¹¹⁹ + 1) 

Trying to factor by splitting the middle term

 1.8     Factoring  l¹¹⁹ + 1 + l²³⁸ 

The first term is,  l¹¹⁹  its coefficient is  1 .
The middle term is,  +1  its coefficient is  1 .
The last term, "the constant", is  +l²³⁸ 

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 by splitting the middle term

 1.9     Factoring  1+l³⁵⁷+l⁷¹⁴ 

The first term is,  1  its coefficient is  1 .
The middle term is,  +l³⁵⁷  its coefficient is  1 .
The last term, "the constant", is  +l⁷¹⁴ 

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 :

  (l119+1)•(-l119+1+l238)•(1-l357+l714)•(l119-1)•(l119+1+l238)•(1+l357+l714)