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Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "h1"   was replaced by   "h^1". 

Rearrange:

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

                c^6*h^606-(c^6*h^1202)=0 

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   c⁶h⁶⁰⁶ - c⁶h¹²⁰²  =   -c⁶h⁶⁰⁶ • (h⁵⁹⁶ - 1) 

Trying to factor as a Difference of Squares :

 2.2      Factoring:  h⁵⁹⁶ - 1 

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 : 1 is the square of 1
Check :  h⁵⁹⁶  is the square of  h²⁹⁸ 

Factorization is :       (h²⁹⁸ + 1)  •  (h²⁹⁸ - 1) 

Trying to factor as a Difference of Squares :

 2.3      Factoring:  h²⁹⁸ - 1 

Check : 1 is the square of 1
Check :  h²⁹⁸  is the square of  h¹⁴⁹ 

Factorization is :       (h¹⁴⁹ + 1)  •  (h¹⁴⁹ - 1) 

Equation at the end of step  2  :

  -c6h606 • (h298 + 1) • (h149 + 1) • (h149 - 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   -c⁶h⁶⁰⁶  = 0

Setting any of the variables to zero solves the equation:

 c  =  0
 h  =  0

Solving a Single Variable Equation :

 3.3      Solve  :    h²⁹⁸+1 = 0 

 Subtract  1  from both sides of the equation : 
                      h²⁹⁸ = -1
                     h  =  298th root of (-1) 

 The equation has no real solutions. It has 298 imaginary, or complex solutions.
 These solutions are h = 298th root of -1.00000

Solving a Single Variable Equation :

 3.4      Solve  :    h¹⁴⁹+1 = 0 

 Subtract  1  from both sides of the equation : 
                      h¹⁴⁹ = -1
                     h  =  149th root of (-1) 

 Negative numbers have real 149th roots.
 149th root of (-1) = ¹⁴⁹√ -1• 1  = ¹⁴⁹√ -1 • ¹⁴⁹√ 1  =(-1)•¹⁴⁹√ 1 

The equation has one real solution, a negative number This solution is  h = negative

Solving a Single Variable Equation :

 3.5      Solve  :    h¹⁴⁹-1 = 0 

 Add  1  to both sides of the equation : 
                      h¹⁴⁹ = 1
                     h  =  149th root of (1) 

 The equation has one real solution
This solution is  h =

1.  h =
2.  h = negative
3.  These solutions are h = 298th root of -1.00000
4.  h  =  0
5.  c  =  0