Re-Solving Equation

On Breaking-down any Triangle into
a pair of Mirror-Imaged Equilateral Triangles

Section XLIV - Solving the 3rd-Order Polynomial Equation again in Part VII

In this Section , we shall proceed to solve the { 3rd-Order Polynomial Equation } .

In essence , this is a repeat of Section XXI in Part IV of the paper ;
- but with a new twist in this Part VII here .

Let us bring in again the { 3rd-Order Polynomial Equation } from the last Section - Section XLIII :
- i.e. this equation :
Let us now reformat this equation for our further analysis below ,
- by introducing the fixed values [ M ] and [ N ] , as follows :
And the new format of the equation is then :
And we take special note here that :
- Once the complex values [ A ] , [ B ] and [ C ] are fixed ,
  - the complex values [ K1 ] , [ KO ] , [ M ] and [ N ] will also be fixed accordingly .

Let us now bring in an Algebriac Identity involving the Complex Variabes [ T ] and [ U ] .
- We then have this expression for the { cube of ( [ T ] + [ U ] ) } :
  Consequently :
  And on re-arranging terms , we have :
This then is the Algebraic Identity we shall be using below .

Let us now bring back these 2 equations from above for a quick comparison :
- First , the reformatted { 3rd-Order Polynomial Equation } :
- Secondly , the Algebraic Identity :
And we take note here that :
- IF :
  - these 2 particular conditions are fully met , i.e. :
    - Condition ONE :
    - Condition TWO :
  THEN :
  - for the particular set of { [ T ] and [ U ] } satisfying the 2 conditions ,
    - we can always set up a corresponding value of [ x ] given by :
      and consequently yielding :
  SO THAT :
  - the 2 equations above will match exactly numerical-wise :
    - for the particular set of values for { [ T ] , [ U ] and [ x ] } .
We shall now proceed to find a set of values of [ T ] and [ U ] :
- that satisfy both { Condition ONE } and { Condition TWO } ,
so that :
- a solution value of [ x ] can be obtained for the reformatted { 3rd-Order Polynomial Equation } via :

Let us now set up the fixed complex values [ PHI-1 ] and [ PHI-2 ] as the 2 distinct-and-separate square-roots of [ M ] .

We then have the following relations :
Consequently we also have :
Let us now randomly select either [ PHI-1 ] or [ PHI-2 ] as the { pivotal base value }
- for the demostration of our new solution procedure to follow .
And we shall now select [ PHI-1 ] .

Suffice to say at this point that :
- had we selected [ PHI-2 ] instead , the entire new solution procedure to follow would be the same .

Now that we have selected [ PHI-1 ] as the { pivotal base value } for our solution procedure here ,
- we can set up the complex variables [ Tau ] and [ Mu ] based on complex variables [ T ] and [ U ] :
Consequently , we have the relations :
Let us bring back { Condtion ONE } from above :
Substituting therein the values of [ T ] and [ U ] just established immediately preceeding , we have :
Taking note here that :
- the above equation would then become :
Consequently , we have :
- yielding :
And this relation above shall be known as the { NEW Condition ONE } from here-on-in in this Part VII of the paper .
Let us now set up the fixed-value [ H ] based on [ N ] via :
Consequently , we have :
Let us now bring in { Condition TWO } from above :
And based on :
- these relations from above , i.e. :
  and from immediately above :
we can now re-write the above { Condition TWO } as :
- yielding :
And this relation above shall be known as the { NEW Condition TWO } from here-on-in in this Part VII of the paper .