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.