example of solving a cubic equation

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

x3+3x2-1= 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

y3+3y-2= 0. (2)

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

(u3+v3-2)+3(uv+1)(u+v)= 0,

one can determine u and v such that  u3+v3-2=0  and  uv+1=0,  i.e.

{u3+v3= 2,u3v3=-1.

Using the properties of quadratic equation, we infer that u3 and v3 are the roots of the resolvent equation

z2-2z-1= 0.

Therefore, u and v satisfy the binomial equations

u3= 1+2,v3= 1-2, (3)

respectively.  If we choose the real radicalsMathworldPlanetmathPlanetmathu=u0=1+23  and  v=v0=1-23,  the other solutions u,v of (3) are

ζu0,ζ2v0;ζ2u0,ζv0, (4)

where  ζ=-1+i32,ζ2=-1-i32  are the primitive third roots of unityMathworldPlanetmath.  One must combine the pairs  (u,v)  of (4) so that


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


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 partsDlmfMathworldPlanetmath, we have the following solution of (1):


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