example of solving a cubic equation

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

$x^3+3x^2-1=\mathrm{\hspace{0.33em}0}.$ (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

$y^3+3y-2=\mathrm{\hspace{0.33em}0}.$ (2)

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

$$(u^3+v^3-2)+3(uv+1)(u+v)=\mathrm{\hspace{0.33em}0},$$

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

$\{\begin{array}{cc}{u}^3+v^3=\mathrm{\hspace{0.33em}2},\hfill & \\ u^3{v}^3=-1.\hfill & \end{array}$

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

$$z^2-2z-1=\mathrm{\hspace{0.33em}0}.$$

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

$u^3=\mathrm{\hspace{0.33em}1}+\sqrt{2},v^3=\mathrm{\hspace{0.33em}1}-\sqrt{2},$ (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

$\zeta u_0,{\zeta }^2{v}_0;{\zeta }^2{u}_0,\zeta v_0,$ (4)

where  $\zeta =\frac{-1+i\sqrt{3}}{2},{\zeta }^2=\frac{-1-i\sqrt{3}}{2}$  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

$\{\begin{array}{cc}{y}_1=u_0+v_0=\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}},\hfill & \\ y_2=\zeta u_0+{\zeta }^2{v}_0=\frac{-1+i\sqrt{3}}{2}\sqrt[3]{1+\sqrt{2}}+\frac{-1-i\sqrt{3}}{2}\sqrt[3]{1-\sqrt{2}},\hfill & \\ y_3={\zeta }^2{u}_0+\zeta v_0=\frac{-1-i\sqrt{3}}{2}\sqrt[3]{1+\sqrt{2}}+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{1-\sqrt{2}}.\hfill & \end{array}$

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):

$\{\begin{array}{cc}{x}_1=-1+\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}},\hfill & \\ x_2=-1-\frac{1}{2}\left(\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}\right)+i\frac{\sqrt{3}}{2}\left(\sqrt[3]{1+\sqrt{2}}-\sqrt[3]{1-\sqrt{2}}\right),\hfill & \\ x_3=-1-\frac{1}{2}\left(\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}\right)-i\frac{\sqrt{3}}{2}\left(\sqrt[3]{1+\sqrt{2}}-\sqrt[3]{1-\sqrt{2}}\right).\hfill & \end{array}$

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
Canonical name	ExampleOfSolvingACubicEquation
Date of creation	2015-11-18 14:15:18
Last modified on	2015-11-18 14:15:18
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	14
Author	pahio (2872)
Entry type	Example
Classification	msc 12D10
Synonym	example of using Cardano’s formulas
Related topic	PolynomialEquationOfOddDegree
Related topic	ConjugatedRootsOfEquation2