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