- 2
*x*^{3}- 4*x*^{2}+ 3*x*- 4 = 0

*a*_{3}*x*^{3}+*a*_{2}*x*^{2}+*a*_{1}*x*+*a*_{0}= 0

Solving a cubic equation amounts to finding the rootss of a cubic function.
Every cubic equation has at least one solution *x* among the real numbers. The following qualitatively different cases are possible:

- Three different real solutions
- Two real solutions, one of them is a double solution
- A single real solution which is a triple solution
- A single real solution and a pair of complex conjugate solutions which are complex numbers.

The solutions can be found with the following method due to Tartaglia and published by Gerolamo Cardano in 1545.

We first divide the given equation by *a*_{3} to arrive at an equation of the form

*x*^{3}+*ax*^{2}+*bx*+*c*= 0

*t*^{3}+*pt*+*q*= 0. (1)

*u*^{3}-*v*^{3}=*q*-
*uv*=*p*/3

*t*=*v*-*u*

The above system for *u* and *v* can always be solved: solve the second equation for *v*, substitute into the first equation, solve the resulting quadratic equation for *u*^{3}, then take the cube root to find *u*. In some cases the quadratic equation will give complex solutions, even though at least one solution *t* of (1) will be real. This was already noticed by Cardano and is a strong argument for the usefulness (if not the existence) of complex numbers.

Once the values for *t* are known, the substitution *x* = *t* - *a*/3 can be undone to find the values of *x* solving the original equation.

So, if we have an equation

- and

- and

Note that in finding *u*, there were 6 possibilities, since there are two solutions to the square root, and three complex solutions to the cubic root. However, which solution to the square root is chosen does not affect the final resulting *x*.

*See also:* linear equation, quadratic equation, quartic equation, quintic equation, Omar Khayyam

*To do:*

- Example