Solution - Equations reducible to quadratic form

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((x6) -  (23•53x3)) +  8000  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  x⁶-1000x³+8000 

The first term is,  x⁶  its coefficient is  1 .
The middle term is,  -1000x³  its coefficient is  -1000 .
The last term, "the constant", is  +8000 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 8000 = 8000 

Step-2 : Find two factors of  8000  whose sum equals the coefficient of the middle term, which is   -1000 .

For tidiness, printing of 50 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  2  :

  x6 - 1000x3 + 8000  = 0 

Step  3  :

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 3.1     Solve   x⁶-1000x³+8000 = 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  w , such that  w = x³  transforms the equation into :
 w²-1000w+8000 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   991.9350  or   8.0650

Now that we know the value(s) of  w , we can calculate  x  since  x  is  ∛ w  

Doing just this we discover that the solutions of
   x⁶-1000x³+8000 = 0
  are either : 
     x = ∛991.935 = 9.9730
   or:
   x = ∛ 8.065 = 2.0054

Two solutions were found :

1.  x = ∛ 8.065 = 2.0054
2.    x = ∛991.935 = 9.9730