# example of solving a cubic equation

Let us use Cardano’s formulae for solving algebraically the cubic equation

 $\displaystyle x^{3}\!+\!3x^{2}\!-\!1\;=\;0.$ (1)

First apply the Tschirnhaus transformation (http://planetmath.org/CardanosDerivationOfTheCubicFormula)  $x:=y\!-\!1$  for removing the quadratic term; from  $(y-1)^{3}+3(y-1)^{2}-1=0$  we get the simplified equation

 $\displaystyle y^{3}\!+\!3y\!-\!2\;=\;0.$ (2)

We now suppose that  $y:=u\!+\!v$.  Substituting this into (2) and rewriting the equation in the form

 $(u^{3}\!+\!v^{3}\!-\!2)+3(uv\!+\!1)(u\!+\!v)\;=\;0,$

one can determine $u$ and $v$ such that  $u^{3}\!+\!v^{3}\!-\!2=0$  and  $uv\!+\!1=0$,  i.e.

 $\displaystyle\begin{cases}u^{3}\!+\!v^{3}\;=\;2,\\ u^{3}v^{3}\;=\;-1.\end{cases}$

Using the properties of quadratic equation, we infer that $u^{3}$ and $v^{3}$ are the roots of the resolvent equation

 $z^{2}\!-\!2z\!-\!1\;=\;0.$

Therefore, $u$ and $v$ satisfy the binomial equations

 $\displaystyle u^{3}\;=\;1\!+\!\sqrt{2},\qquad v^{3}\;=\;1\!-\!\sqrt{2},$ (3)

respectively.  If we choose the real radicals$u=u_{0}=\sqrt[3]{1\!+\!\sqrt{2}}$  and  $v=v_{0}=\sqrt[3]{1\!-\!\sqrt{2}}$,  the other solutions $u,\,v$ of (3) are

 $\displaystyle\zeta u_{0},\;\;\zeta^{2}v_{0};\qquad\zeta^{2}u_{0},\;\;\zeta v_{% 0},$ (4)

where  $\zeta=\frac{-1+i\sqrt{3}}{2},\quad\zeta^{2}=\frac{-1-i\sqrt{3}}{2}$  are the primitive third roots of unity.  One must combine the pairs  $(u,\,v)$  of (4) so that

 $uv\;=\;\sqrt[3]{u^{3}v^{3}}\;=\;-1.$

Accordingly, all three roots of the cubic equation (2) are

 $\displaystyle\begin{cases}y_{1}\;=\;u_{0}\!+\!v_{0}\;=\;\sqrt[3]{1\!+\!\sqrt{2% }}\!+\!\sqrt[3]{1\!-\!\sqrt{2}},\\ y_{2}\;=\;\zeta u_{0}\!+\!\zeta^{2}v_{0}\;=\;\frac{-1+i\sqrt{3}}{2}\sqrt[3]{1% \!+\!\sqrt{2}}\!+\!\frac{-1-i\sqrt{3}}{2}\sqrt[3]{1\!-\!\sqrt{2}},\\ y_{3}\;=\;\zeta^{2}u_{0}\!+\!\zeta v_{0}\;=\;\frac{-1-i\sqrt{3}}{2}\sqrt[3]{1% \!+\!\sqrt{2}}\!+\!\frac{-1+i\sqrt{3}}{2}\sqrt[3]{1\!-\!\sqrt{2}}.\\ \end{cases}$

The roots of the original equation (1) are gotten via the used substitution equation  $x:=y-1$, i.e. adding $-1$ to the values of $y$.  If we also separate the real (http://planetmath.org/RealPart) and imaginary parts, we have the following solution of (1):

 $\displaystyle\begin{cases}x_{1}\;=\;-1\!+\!\sqrt[3]{1\!+\!\sqrt{2}}\!+\!\sqrt[% 3]{1\!-\!\sqrt{2}},\\ x_{2}\;=\;-1\!-\!\frac{1}{2}\!\left(\sqrt[3]{1\!+\!\sqrt{2}}\!+\!\sqrt[3]{1\!-% \!\sqrt{2}}\right)\!+i\frac{\sqrt{3}}{2}\!\left(\sqrt[3]{1\!+\!\sqrt{2}}\!-\!% \sqrt[3]{1\!-\!\sqrt{2}}\right),\\ x_{3}\;=\;-1\!-\!\frac{1}{2}\!\left(\sqrt[3]{1\!+\!\sqrt{2}}\!+\!\sqrt[3]{1\!-% \!\sqrt{2}}\right)\!-i\frac{\sqrt{3}}{2}\!\left(\sqrt[3]{1\!+\!\sqrt{2}}\!-\!% \sqrt[3]{1\!-\!\sqrt{2}}\right).\end{cases}$

One of the roots is a real number, but the other two are (i.e. non-real) complex conjugates of each other.

Title example of solving a cubic equation ExampleOfSolvingACubicEquation 2015-11-18 14:15:18 2015-11-18 14:15:18 pahio (2872) pahio (2872) 14 pahio (2872) Example msc 12D10 example of using Cardano’s formulas PolynomialEquationOfOddDegree ConjugatedRootsOfEquation2