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Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "c1"   was replaced by   "c^1".  1 more similar replacement(s).

Rearrange:

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

                     c^12*c*c-(c^119)=0 

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   c¹⁴ - c¹¹⁹  =   -c¹⁴ • (c¹⁰⁵ - 1) 

Trying to factor as a Difference of Cubes:

 2.2      Factoring:  c¹⁰⁵ - 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 :  c¹⁰⁵ is the cube of   c³⁵

Factorization is :
             (c³⁵ - 1)  •  (c⁷⁰ + c³⁵ + 1) 

Trying to factor by splitting the middle term

 2.3     Factoring  c⁷⁰ + c³⁵ + 1 

The first term is,  c⁷⁰  its coefficient is  1 .
The middle term is,  +c³⁵  its coefficient is  1 .
The last term, "the constant", is  +1 

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

Equation at the end of step  2  :

  -c14 • (c35 - 1) • (c70 + c35 + 1)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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.

Solving a Single Variable Equation :

 3.2      Solve  :    -c¹⁴ = 0 

 Multiply both sides of the equation by (-1) :  c¹⁴ = 0


                     c  =  14th root of (0) 

 Any root of zero is zero. This equation has one solution which is  c = 0

Solving a Single Variable Equation :

 3.3      Solve  :    c³⁵-1 = 0 

 Add  1  to both sides of the equation : 
                      c³⁵ = 1
                     c  =  35th root of (1) 

 The equation has one real solution
This solution is  c =

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 3.4     Solve   c⁷⁰+c³⁵+1 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  b , such that  b = c³⁵  transforms the equation into :
 b²+b+1 = 0

Solving this new equation using the quadratic formula we get two imaginary solutions :
   b = -0.5000 ± 0.8660 i 
Now that we know the value(s) of  b , we can calculate  c  since  c  is the 35 root of   b  

Since we are speaking 35th root, each of the two imaginary solutions of has 35 roots

Tiger finds these roots using de Moivre's Formula

The 35th roots of  -0.500 + 0.866 i   are:

35th roots of  -0.500- 0.866 i  :
 

1.  c =
2.  c = 0