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[parent] example of solving a cubic equation (Example)

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 $ 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
$\displaystyle (u^3+v^3-2)+3(uv+1)(u+v) = 0,$
one can determine $ u$ and $ v$ such that $ u^3+v^3-2 = 0$ and $ uv+1 = 0$, i.e.
\begin{align*}\begin{cases}u^3+v^3 = 2,\\ u^3v^3 = -1. \end{cases}\end{align*}    

Using the properties of quadratic equation, we infer that $ u^3$ and $ v^3$ are the roots of the resolvent equation
$\displaystyle z^2-2z-1 = 0.$
Therefore, $ u$ and $ v$ satisfy the binomial equations
$\displaystyle u^3 = 1+\sqrt{2}, \quad v^3 = 1-\sqrt{2},$ (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
$\displaystyle \zeta u_0,\;\;\zeta^2u_0; \quad \zeta v_0,\;\;\zeta^2v_0,$ (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
$\displaystyle uv = \sqrt[3]{u^3v^3} = -1.$
Accordingly, all three roots of the cubic equation (2) are
\begin{align*}\begin{cases}y_1 = u_0+v_0 = \sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt... ...+\sqrt{2}}+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{1-\sqrt{2}}.\\ \end{cases}\end{align*}    

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 and imaginary parts, we have the following solution of (1):
\begin{align*}\begin{cases}x_1 = -1+\sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}},\\... ...}\left(\sqrt[3]{1+\sqrt{2}}-\sqrt[3]{1-\sqrt{2}}\right). \end{cases}\end{align*}    

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



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See Also: polynomial equation of odd degree, conjugated roots of equation

Other names:  example of using Cardano's formulas

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Cross-references: complex conjugates, imaginary parts, roots of unity, primitive, solutions, radicals, real, binomial equations, resolvent equation, roots, properties of quadratic equation, equation, term, cubic equation, Cardano's formulae
There are 2 references to this entry.

This is version 9 of example of solving a cubic equation, born on 2008-02-26, modified 2008-02-27.
Object id is 10335, canonical name is ExampleOfSolvingACubicEquation.
Accessed 898 times total.

Classification:
AMS MSC: 12D10 (Field theory and polynomials :: Real and complex fields :: Polynomials: location of zeros )
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