User:MathMan64/CasusIrreducibilis

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[edit] Casus

Some cubic equations with real solutions have solutions that can not be written exactly, without the use of imaginary expressions.

I plan to find out how to tell which a cubic equations have solutions that are totally real expressions, and to relate this topic to the exact value of trigonometric functions of angles whose degree measure is not a multiple of three.

Criteria for irreducibile polynomials are given by Eisenstein's criterion

[edit] Solving Cubic Equations

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

Let y=x+\frac a 3 \ \

This yields y^3+Py+Q=0 \

Where P=\frac{3b-a^2} 3

and Q=\frac{27c-9ab+2a^3} {27} \

Cardano's Method gives:

y=\sqrt [3] {- \frac Q 2 + \sqrt{ \frac{P^3}{27} \ + \ \frac{Q^2} 4 }} + \sqrt [3] { - \frac Q 2 - \sqrt{ \frac{P^3}{27} + \frac{Q^2} 4 } }

Which can be expanded by assigning:

H= \frac Q 2 =\frac {27c-9ab+2a^3}{54}

and G = \frac{Q^2} 4 = \frac {4a^6-36a^4b+108a^3c+81a^2b^2-486abc+729c^2} {2916}

and F=\frac{P^3}{27} =\frac{-a^6+9a^4b-27a^2b^2+27b^3}{729}

and K=F+G= \frac{4a^3c-a^2b^2-18abc+4b^3+27c^2}{108}

and \sqrt K =\frac \sqrt 3 {18} \sqrt{4a^3c-a^2b-18abc+4b^3+27c^2}

So simply expressed: y=\sqrt [3]{(-H+ \sqrt K)} + \sqrt [3]{(-H- \sqrt K}

and x=\sqrt [3]{(-H+ \sqrt K)} + \sqrt [3]{(-H- \sqrt K)} -\frac a 3

Or more monstrously expressed:

x=\frac{ \sqrt[3]{4(9ab-2a^3-27c+ 3\sqrt{4a^3c-a^2b^2-18abc+4b^3+27c^2} )}}6 +

\frac{ \sqrt[3]{4(9ab-2a^3-27c- 3\sqrt{4a^3c-a^2b^2-18abc+4b^3+27c^2} )}}6 - \frac a 3
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