It is actually very simple. To solve an equation of the form:
x^3 + a x^2 + b x + c = 0
you follow the following steps.
First, get rid of the quadratic term
a x^2 using the substitution:
x = y - a/3
This step is analogous to how you solve the quadratic equation when you write it as a perfect square.
In this case we aren't finshed yet as the equation now is of the form:
y^3 + p y + q = 0
How do we solve this equation?
Consider the identity:
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
We can rewrite the right hand side as follows:
a^3 + 3a^2b + 3ab^2 + b^3 =
a^3 + b^3 + 3ab(a+b)
So, we have:
(a+b)^3 = 3ab(a+b)+ a^3 + b^3 ---->
(a+b)^3 - 3ab(a+b) - [a^3+b^3] = 0
Of course, this equation is always satisfied whatever values for a and b you substitute in it. Now, if we choose a and b such that:
3ab = -p (1)
a^3 + b^3 = -q (2)
Then:
(a+b)^3 +p(a+b) + q = 0
so y = a + b is then the solution we are looking for.
So, it all boils down to solving equations (1) and (2) for a and b.
If you thake the third power of (1):
27 a^3b^3 = -p^3
So, if we put A = a^3 and B = B^3, we have:
A*B = -p^3/27
A + B = -q
If you eliminate, say, B , you get a quadratic equation for A which you know how to solve. You then only need to pay attention when you extract the cube roots to find a and b. Equation (1) has to be satisfied, which means that if you choose one of the three possible complex cube roots for A, the root for B is fixed.