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Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x8"   was replaced by   "x^8". 

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

           4/(x^8)-3/(x-8)-((3*x)/(x^(2)-64))=0 

Step  1  :

               3x  
 Simplify   ———————
            x2 - 64

Trying to factor as a Difference of Squares :

 1.1      Factoring:  x² - 64 

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 : 64 is the square of 8
Check :  x²  is the square of  x¹ 

Factorization is :       (x + 8)  •  (x - 8) 

Equation at the end of step  1  :

     4    3         3x    
  (————-—————)-———————————  = 0 
   (x8) (x-8)  (x+8)•(x-8)

Step  2  :

              3  
 Simplify   —————
            x - 8

Equation at the end of step  2  :

     4   3        3x    
  (————-———)-———————————  = 0 
   (x8) x-8  (x+8)•(x-8)
 Step  3  :
             4
 Simplify   ——
            x8

Equation at the end of step  3  :

    4      3               3x       
  (—— -  —————) -  —————————————————  = 0 
   x8    x - 8     (x + 8) • (x - 8)

Step  4  :

Calculating the Least Common Multiple :

 4.1    Find the Least Common Multiple

      The left denominator is :       x⁸ 

      The right denominator is :       x-8 

      Least Common Multiple:
      x⁸ • (x-8) 

Calculating Multipliers :

 4.2    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = x-8

   Right_M = L.C.M / R_Deno = x⁸

Making Equivalent Fractions :

 4.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)²   and  (y²+y)/(y+1)³  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

   L. Mult. • L. Num.       4 • (x-8)
   ——————————————————  =   ——————————
         L.C.M             x8 • (x-8)

   R. Mult. • R. Num.        3 • x8  
   ——————————————————  =   ——————————
         L.C.M             x8 • (x-8)

Adding fractions that have a common denominator :

 4.4       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 4 • (x-8) - (3 • x8)     -3x8 + 4x - 32
 ————————————————————  =  ——————————————
      x8 • (x-8)           x8 • (x - 8) 

Equation at the end of step  4  :

  (-3x8 + 4x - 32)            3x       
  ———————————————— -  —————————————————  = 0 
    x8 • (x - 8)      (x + 8) • (x - 8)

Step  5  :

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   -3x⁸ + 4x - 32  =   -1 • (3x⁸ - 4x + 32) 

Polynomial Roots Calculator :

 6.2    Find roots (zeroes) of :       F(x) = 3x⁸ - 4x + 32
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  3  and the Trailing Constant is  32.

 The factor(s) are:

of the Leading Coefficient :  1,3
 of the Trailing Constant :  1 ,2 ,4 ,8 ,16 ,32

 Let us test ....

Note - For tidiness, printing of 19 checks which found no root was suppressed

Polynomial Roots Calculator found no rational roots

Calculating the Least Common Multiple :

 6.3    Find the Least Common Multiple

      The left denominator is :       x⁸ • (x - 8) 

      The right denominator is :       (x + 8) • (x - 8) 

      Least Common Multiple:
      x⁸ • (x - 8) • (x + 8) 

Calculating Multipliers :

 6.4    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = x + 8

   Right_M = L.C.M / R_Deno = x⁸

Making Equivalent Fractions :

 6.5      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      (-3x8+4x-32) • (x+8)
   ——————————————————  =   ————————————————————
         L.C.M              x8 • (x-8) • (x+8) 

   R. Mult. • R. Num.            3x • x8     
   ——————————————————  =   ——————————————————
         L.C.M             x8 • (x-8) • (x+8)

Adding fractions that have a common denominator :

 6.6       Adding up the two equivalent fractions

 (-3x8+4x-32) • (x+8) - (3x • x8)     -6x9 - 24x8 + 4x2 - 256
 ————————————————————————————————  =  ———————————————————————
        x8 • (x-8) • (x+8)            x8 • (x - 8) • (x + 8) 

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   -6x⁹ - 24x⁸ + 4x² - 256  = 

  -2 • (3x⁹ + 12x⁸ - 2x² + 128) 

Checking for a perfect cube :

 7.2    3x⁹ + 12x⁸ - 2x² + 128  is not a perfect cube

Trying to factor by pulling out :

 7.3      Factoring:  3x⁹ + 12x⁸ - 2x² + 128 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -2x² + 128 
Group 2:  12x⁸ + 3x⁹ 

Pull out from each group separately :

Group 1:   (x² - 64) • (-2)
Group 2:   (x + 4) • (3x⁸)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

 7.4    Find roots (zeroes) of :       F(x) = 3x⁹ + 12x⁸ - 2x² + 128

     See theory in step 6.2
In this case, the Leading Coefficient is  3  and the Trailing Constant is  128.

 The factor(s) are:

of the Leading Coefficient :  1,3
 of the Trailing Constant :  1 ,2 ,4 ,8 ,16 ,32 ,64 ,128

 Let us test ....

Note - For tidiness, printing of 27 checks which found no root was suppressed

Polynomial Roots Calculator found no rational roots

Equation at the end of step  7  :

  -2 • (3x9 + 12x8 - 2x2 + 128)
  —————————————————————————————  = 0 
     x8 • (x - 8) • (x + 8)    

Step  8  :

When a fraction equals zero :

 8.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  -2•(3x9+12x8-2x2+128)
  ————————————————————— • x8•(x-8)•(x+8) = 0 • x8•(x-8)•(x+8)
     x8•(x-8)•(x+8)    

Now, on the left hand side, the  x⁸ • x-8 • x+8  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -2  •  (3x⁹+12x⁸-2x²+128)  = 0

Equations which are never true :

 8.2      Solve :    -2   =  0

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

Equations of order 5 or higher :

 8.3     Solve   3x⁹+12x⁸-2x²+128 = 0

Points regarding equations of degree five or higher.

 (1)  There is no general method (Formula) for solving polynomial equations of degree five or higher.

 (2)  By the Fundamental theorem of Algebra, if we allow complex numbers, an equation of degree  n  will have exactly  n  solutions
(This is if we count double solutions as  2 , triple solutions as  3  and so on

) (3)  By the Abel-Ruffini theorem, the solutions can not always be presented in the conventional way using only a finite amount of additions, subtractions, multiplications, divisions or root extractions

 (4)  If  F(x)  is a polynomial of odd degree with real coefficients, then the equation  F(X)=0  has at least one real solution.

 (5)  Using methods such as the  Bisection  Method, real solutions can be approximated to any desired degree of accuracy. Failed to find the initial interval for implementing the BiSection Method

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