301 Abstract Algebra
2:00 - 2:50 MWF
Hentzel 432 Carver
hentzel@iastate.edu
Web Page   http://www.math.iastate.edu/hentzel/class.301

Text: A First Course in Abstract Algebra, sixth Edition
by John B. Fraleigh
Course: Chapter 1,2,3,4,5

Friday, November 30.  

Main Idea:  How to solve the cubic.

Key Words:  Cardan's Formula, Cubic Formula, 

Goal:  Learn the steps to solve a cubic, but not to do so

by substituting in a formula.


The quadratic formula gives the solution for

                                     -b +/- Sqrt[b^2-4ac]
ax^2 + bx + c = 0.  The solution is  -------------------- 
                                            2a

The formula comes from completing the square.


       ax^2 + bx + c = 0

       x^2 + b/a x   =  -c/a

       x^2 + b/a x +(b/2a)^2  = -c/a +(b/2a)^2

                       b^2 -4ac
        (x+b/2a)^2 =  ---------------
                         4a^2

                      +/- Sqrt[b^2-4ac]
         x+b/2a  =   --------------------
                           2a
                   
                    -b +/- Sqrt[b^2-4ac]
          x    =  ------------------------
                            2a

The solution to

       x^3 + ax + b = 0


substitute x = z+r/z


     (z+r/z)^3 + a(z+r/z) + b  = 0


       z^3 + 3 z^2 (r/z) + 3 z (r/z)^2 + (r/z)^3 + az + ar/z + b = 0


       z^3 + z(3r + a) + (r/z)(a+3r) +(r/z)^3 + a + b = 0

Let r = -a/3

       z^3   -(a^3 /27)/z^3 + b = 0


       (z^3)^2 + b z^3 - a^3 /27 = 0

               -b +/- Sqrt[ b^2 - 4 a^3 /27]
        z^3 = --------------------------------
                         2

                      -b +/- Sqrt[ b^2 - 4 a^3 /27]
        z = CubeRoot[------------------------------ ] 
                                2

Now you can get the x in the original equation.

         x = z-(a/3)/z


Solve   x^3 -   3x  + 1 = 0

Let x = z+1/z.

       (z+1/z)^3 - 3(z+1/z) + 1 = 0


        z^3 + 1 + 1/z^3 = 0


        (z^3)^2 + z^3 + 1 = 0

             -1 +/- i Sqrt[3]
      z^3 = -----------------
                  2


      z^3 = Cos[120]+I Sin[120]

      z  = Cos[40]+ I Sin[40]


      x = Cis[40] + Cis[-40]  = 2 Cos[40]


                         2 Pi          2 Pi 3
f[2 Cos[40] =  1 - 6 Cos[----] + 8 Cos[----]
                          9             9


Cos[2 Pi/9] = 0.766044



Hand-in-homework

1.  Solve  4 x^3 - 3x = 1/Sqrt[2] 



2.  Solve  4 x^3 - 3x = Sqrt[3]/2


3.  Find a substitution x = y+r so that the equation

 x^3 + ax^2 + bx + c = 0

 loses the x^2 term to become


   y^3 + b' y + c'  = 0

so that then the previous technique can be used.