An example of a cubic equation with 3 real roots that can be solved using the cubic formula without imaginary?

(x-1)(x-2)(x-3) is a cubic equation with roots 1,2 and 3
if it is a cubic equation it si not a must that it shoul have imaginary roots
this would (x-1)(x^2+1)

Is it possible? I'm sure it is. But as you can see the cubic formula is really nasty and hard to work with. If your trying to solve a cubic that happens to have 3 real solutions then odds are it will be easier to simplify the equation and solve, than to use the cubic.

The answer is actually no; it is not possible. This is known as the causis irreducibilis and is one of the reasons that the cubic formula is not often used. Another is that even very simply roots such as 2,3, and 5 can look very strange when the cubic formula is used.

The easiest way to see this is to assume you have such a cubic. Usually the cubic equation is done for cubics of the form
x^3 +cx=d. Since every cubic can be transformed into one of this form, this is not a difficulty.

For this equation to have three real roots c must be negative and d must be between the relative maximum and the relative minimum values of the polynomial x^3+bx. When you write down this condition, it is exactly the same as the condition that the square root in the cubic formula be imaginary. If you want more details, just write.