Fermat's Last Theorem- applications of Prime factors method

Fermat's Last Theorem- Applications
If x is given a whole integer. Then
x3 = z3 - y3  = (z - y) (z2 + zy + y2)
x(odd) = P01P02  and x3  = P013P023
Since x+y > z , x > z - y
 The distribution of the prime factors of x3 among its two product factors should satisfy this condition.
z - y = P01  then z2 + zy + y2 =P012P023
z= P01 + y substituting this value of z in the other  equation
z2 + zy + y2 = 3y2 + 3y P01 +  P012 = P012P023
3y2 + 3y P01 +  P012 - P012P023 = 0
This is a quadratic equation in terms of y, solving for y we get,
y = [ -3P01 +  √  9 P012 -12 (P012 - P012P023)]  / 6
y = [ -3P01 + √  12 P012P023 - 3P012 ]  /6
If x = P01P02  = 3x7 , then
y = [-3 + √ 4113]  / 2 and c = [3 +  √ 4113] / 2

The conservation of odd-evenness of the cubical relations demands that z - y = 23n where n is a number.
when n=1 z - y= 8 and y is singly even,when n = 2 x - y = 64 and y is doubly even and so on.
y 3  = 23n P013 ........... (any number of odd prime factors)
y3 = z3 - x3 = (z - x) (z2 +zx + x2)
where z - x is even where as z2 +zx + x2 is odd, hence z - x = 23n or one or more prime factors may be attached with it.. All the remaining prime factors multiplied together  must be equal to z2 +zx + x2 . For example
 y = 2P01  then    y3 = 23P013 = z3 - x3  = (z - x) (z2 +zx + x2)
z - x  = 8 or z = x + 8. Substituting this value of z in z2 +zx + x2 = P013
3a2 +24a +64 - P013  = 0
or a =[  -12 + √  3P013  - 48] / 3
If P01  = 7 , x = - 4 +  √109  and  z = 4 + √  109
If P01  = 11, x = [ -12 + √3945]/3 and z = x+ 8  = [ 12 + √ 3945]/3