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Solving a cubic (or rather simplifying a real root in complex form)

  1. Oct 20, 2005 #1
    The determinate of the following 3x3 matrix

    1-y, 2 , 3
    2 , 4-y, 5
    3 , 5 , 7-y

    gives a cubic that simplifies to,

    y^3 - 12*y^2 + y + 1 = 0.

    Now, apparently the teacher picked random numbers for the original matrix, making the problem delve into other realms of mathematics. It isn't solvable by factoring methods, but using Cardano's method...

    y = x +4,

    the cubic becomes,

    (x + 4)^3 - 12*(x + 4)^2 + (x + 4) + 1 = 0.

    This simplifies to,

    x^3 - 47*x -123 = 0,

    which is in the form

    x^3 + b*x + c = 0.

    To solve a cubic like this, b and c are just plugged into the following formula,

    x = ( -(c/2) + ((c/2)^1/2 + (b/3)^1/3)^1/2)^1/3 + (-(c/2) - ((c/2)^1/2 + (b/3)^1/3)^1/2)^1/3,

    which for my cubic comes out...

    x = ((123/2) + (-6809/108)^1/2)^1/3 + ((123/2) - (-6809/108)^1/2)^1/3.

    This is a real root, but I have no idea how to simplify it, ridding it of the complex parts.
    Sorry to show all the lead up, it really wasn't needed for the simplification, but since the rules say to show what you have done... I did.
    Thanks in advance for any help.
    Last edited: Oct 20, 2005
  2. jcsd
  3. Oct 20, 2005 #2


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    Homework Helper

    Thanks for posting your work.

    Cardan's method is illustrated by HallsofIvy in this thread.

    Perhaps it will help you.
  4. Oct 20, 2005 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    There exist real numbers that can only be expressed in terms of radicals by taking a detour through the complex numbers -- it may not be possible to simplify this expression.
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