Let us use Cardano’s formulae for solving algebraically the cubic equation
First apply the Tschirnhaus transformation for removing the quadratic term; from we get the simplified equation
We now suppose that . Substituting this into (2) and rewriting the equation in the form
one can determine and such that and , i.e.
Using the properties of quadratic equation, we infer that and are the roots of the resolvent equation
Therefore, and satisfy the binomial equations
respectively. If we choose the real radicals and , the other solutions of (3) are
where are the primitive third roots of unity. One must combine the pairs 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 , i.e. adding to the values of . If we also separate the real and imaginary parts, we have the following solution of (1):
One of the roots is a real number, but the other two are imaginary (i.e. non-real) complex conjugates of each other.