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Step  1  :

Equation at the end of step  1  :

  (2x215 • x) -  8  = 0 
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x²¹⁶ - 8  =   -2 • (4 - x²¹⁶) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  4 - x²¹⁶ 

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 :  x²¹⁶  is the square of  x¹⁰⁸ 

Factorization is :       (2 + x¹⁰⁸)  •  (2 - x¹⁰⁸) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  2 + x¹⁰⁸ 

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:  2 - x¹⁰⁸ 

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:  2 - x¹⁰⁸ 

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

Equation at the end of step  3  :

  -2 • (x108 + 2) • (2 - x108)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Equations which are never true :

 4.2      Solve :    -2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 4.3      Solve  :    x¹⁰⁸+2 = 0 

 Subtract  2  from both sides of the equation : 
                      x¹⁰⁸ = -2
                     x  =  108th root of (-2) 

 The equation has no real solutions. It has 108 imaginary, or complex solutions.
 These solutions are x = 108th root of -2.00000

Solving a Single Variable Equation :

 4.4      Solve  :    -x¹⁰⁸+2 = 0 

 Subtract  2  from both sides of the equation : 
                      -x¹⁰⁸ = -2
Multiply both sides of the equation by (-1) :  x¹⁰⁸ = 2


                     x  =  108th root of (2) 

 The equation has two real solutions  
 These solutions are  x = ± 108th root of 2 = ± 1.0064  
 

1.  x = ± 108th root of 2 = ± 1.0064
2.  These solutions are x = 108th root of -2.00000