The general form of a polynomial equation of degree 3: takes the form of  (1), when we divide its terms with and set .
Doing the following transformation:

the equation (1) gets the form:

where 

Now doing the transformation:

the equation (2) takes the form:

If the roots of the second degree equation (3) with respect to , then solving one(*) of binomial equations:

we can find three values  for z .

Assigning these values to the transformation (M2) we find three values for y, from which with the help of (M1) we find the roots  of (1)

(*) : it does not matter which one of the two we are going to solve. At the end we will find the same roots
  of the equation (1).

Example

Solve the equation : .

Solution

We must bring the equation to the form (1).

It is : 

Equation (3) becomes: from which we get the following:

and with the help of (M2) we find:

Finally (M1) gives :