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%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Fri Nov 09 16:33:59 2001}
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%$r=-\frac{B}{3A} $
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%H=36,\PARA{038<p type="texpara" tag="Body Text" >College Algebra Fall 2001}
%F=36,\PARA{038<p type="texpara" tag="Body Text" >Al Lehnen (Problem suggested by Andrea Martin)\hfill \thepage }
%}


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\subsubsection{$\protect\vspace{1pt}$Solving the Cubic Equation}

$\vspace{1pt}$Consider the reduced cubic equation: $x^{3}+ax+b=0$

$\vspace{1pt}$By foiling, $(P+Q)^{3}=P^{3}+3P^{2}Q+3PQ^{2}+Q^{3}$ \vspace{1pt%
}so $(P+Q)^{3}-3PQ(P+Q)-(P^{3}+Q^{3})=0$

\vspace{1pt}Let $x=P+Q$ , then $-3PQ=a$ and $-b=P^{3}+Q^{3}$ ; $P=-\frac{a}{%
3Q}$ , so $P^{3}+Q^{3}=-(\frac{a^{3}}{27Q^{3}}-Q^{3})=-b$.

This gives the `Quadratic-Like' equation:

$\frac{a^{3}}{27Q^{3}}-Q^{3}=b$ ; $\frac{a^{3}}{27}-Q^{6}=bQ^{3}$ ; $%
Q^{6}+bQ^{3}-\frac{a^{3}}{27}=0$ . Using the quadratic formula on this
equation gives $Q^{3}=\frac{-b\text{ }\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}}{%
27}}}{2}$ so $Q=\sqrt[3]{\frac{-b\text{ }\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}%
}{27}}}{2}}$ since $P=-\frac{a}{3Q}$ we have $P=\frac{-a}{3\sqrt[3]{\frac{-b%
\text{ }\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}{2}}}=\frac{-a\sqrt[3]{-b%
\text{ }-\text{ }\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}}{3\sqrt[3]{%
\frac{\left( -b\right) ^{2}\text{ }-\text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}%
^{2}}{2}}}$ .

$P=\frac{-a\sqrt[3]{-b\text{ }-\text{ }\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}}{%
27}}}}{3\sqrt[3]{\frac{\left( -b\right) ^{2}\text{ }-\text{ }b^{2}-\frac{%
4a^{3}}{27}}{2}}}=\frac{-a\sqrt[3]{-b\text{ }-\text{ }\pm \text{ }\sqrt{%
b^{2}+\frac{4a^{3}}{27}}}}{3\sqrt[3]{\frac{-\frac{4a^{3}}{27}}{2}}}=\frac{-a%
\sqrt[3]{-b\text{ }-\text{ }\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}}{-3%
\sqrt[3]{2}\frac{a}{3}}=\frac{a\sqrt[3]{-b\text{ }-\text{ }\pm \text{ }\sqrt{%
b^{2}+\frac{4a^{3}}{27}}}}{\sqrt[3]{2}a}=\sqrt[3]{\frac{-b\text{ }-\text{ }%
\pm \text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}{2}}$

Hence, $x=P+Q=\sqrt[3]{\frac{-b\text{ }+\text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}%
}}{2}}+\sqrt[3]{\frac{-b\text{ }-\text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}{2}}%
=-\left( \sqrt[3]{\frac{b-\text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}{2}}+\sqrt[3%
]{\frac{b+\text{ }\sqrt{b^{2}+\frac{4a^{3}}{27}}}{2}}\right) $.

solves $x^{3}+ax+b=0$. \textbf{Note:}this only gives us only\textbf{\ one }%
of three solutions of the reduced cubic equation.

\vspace{1pt}The remaining two solutions need to be found by using the factor
theorem and solving the quadratic equation obtained after factoring out $%
x-(P+Q)$.

The more \textbf{general cubic} equation $Ax^{3}+Bx^{2}+Cx+D=0$ can be
solved by first 

getting into reduced form. Let $x=Z+r$ ; $x^{3}=Z^{3}+3Z^{2}r+3Zr^{2}+r^{3}$
; $x=Z^{2}+2Zr+r^{2}$ so the equation becomes $%
A(Z^{3}+3Z^{2}r+3Zr^{2}+r^{3})+B(Z^{2}+2Zr+r^{2})+C(Z+r)+D=0$

$Z^{3}(A)+Z^{2}(3Ar+B)+Z(3Ar^{2}+2Br+C)+Ar^{3}+Br^{2}+Cr+D=0$ . Now choose $%
r=-\frac{B}{3A}$ i.e., $x=Z-\frac{B}{3A}$ where $Z$ satifies the reduced
cubic $Z^{3}+aZ+b=0$ with $a=\frac{3Ar^{2}+2Br+C}{A}=\allowbreak \frac{%
3CA-B^{2}}{3A^{2}}$ ;

$b=\frac{Ar^{3}+Br^{2}+Cr+D}{A}=\allowbreak \frac{2B^{3}-9ABC+27DA^{2}}{%
27A^{3}}$ . Thus one solution is  $Z=-\left( \sqrt[3]{\frac{b-\text{ }\sqrt{%
b^{2}+\frac{4a^{3}}{27}}}{2}}+\sqrt[3]{\frac{b+\text{ }\sqrt{b^{2}+\frac{%
4a^{3}}{27}}}{2}}\right) $.

$\vspace{1pt}$Consider solving $x^{3}-2x+3=0$ which has $a=-2$ and $b=3$ .

\vspace{1pt}

Thus, $P+Q=-\left( \sqrt[3]{\frac{3-\text{ }\sqrt{9-\frac{32}{27}}}{2}}+%
\sqrt[3]{\frac{3+\text{ }\sqrt{9-\frac{32}{27}}}{2}}\right) =-\left( \sqrt[3%
]{\frac{9-\sqrt{\frac{211}{3}}}{6}}+\sqrt[3]{\frac{9+\sqrt{\frac{211}{3}}}{6}%
}\right) $ which upon rationalizing the dominator gives $\allowbreak P+Q=-%
\frac{1}{6}\sqrt[3]{\left( 324-12\sqrt{633}\right) }-\frac{1}{6}\sqrt[3]{%
\left( 324+12\sqrt{633}\right) }\allowbreak $ . As a decimal approximation
we have $P+Q=\allowbreak -1.\,\allowbreak 893\,289\,196\,\allowbreak
304\,497\,788\,\allowbreak 906\,355\,609\,\allowbreak 7\allowbreak $ \ . 

\vspace{1pt}

The complete solution of $x^{3}-2x+3=0$, Solution is : $x=-\frac{1}{6}\sqrt[3%
]{\left( 324+12\sqrt{633}\right) }-\frac{4}{\sqrt[3]{\left( 324+12\sqrt{633}%
\right) }},\allowbreak x=\frac{1}{12}\sqrt[3]{\left( 324+12\sqrt{633}\right) 
}+\frac{2}{\sqrt[3]{\left( 324+12\sqrt{633}\right) }}+\frac{1}{2}i\sqrt{3}%
\left( -\frac{1}{6}\sqrt[3]{\left( 324+12\sqrt{633}\right) }+\frac{4}{\sqrt[3%
]{\left( 324+12\sqrt{633}\right) }}\right) ,\allowbreak x=\frac{1}{12}\sqrt[3%
]{\left( 324+12\sqrt{633}\right) }+\frac{2}{\sqrt[3]{\left( 324+12\sqrt{633}%
\right) }}-\frac{1}{2}i\sqrt{3}\left( -\frac{1}{6}\sqrt[3]{\left( 324+12%
\sqrt{633}\right) }+\frac{4}{\sqrt[3]{\left( 324+12\sqrt{633}\right) }}%
\right) \allowbreak $

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