How can one solve cubic functions?

[TheDominis]

x³ - 12x + 1 = 0

How does one solve for x?

 

[Mark44]

It's not a trivial process. See this Wikipedia article, http://en.wikipedia.org/wiki/Cubic_function
especially the Summary about halfway down the page.

 

[HallsofIvy]

Let x= a- b. Then $a^3= (a-b)^3= a^3- 3a^2b+ 3ab^2- b^3$.

Also $3abx= 3ab(a- b)= 3a^2b- 3ab^2$.

So $x^3+ 3abx= a^3- b^3$. Letting m= 3ab and $n= a^3- b^3$, then x= a-b satisfies $x^3+ mx= n$.

Suppose we know m and n- can we "recover" a and b and so find x?

If m= 3ab, then b= m/3a and $n= a^3- m^3/3^3a^3$. Multiplying through by $a^3$ we get $na^3= (a^3)^2- m^3/3^3$ which we can think of as a quadratic equation for $a^3$: $(a^3)^2- na^3- m^3/3^3= 0$ and solve by the quadratic formula:
$$a^3= \frac{n\pm\sqrt{n^2+ 4\frac{m^3}{m^3}}}{2}$$$$= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$
so that
$$a= \sqrt[3]{\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}}}$$

Since $a^3- b^3= n$, $b^3= a^3- n$ so
$$b^3= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$
and
$$b= -\sqrt[3]{\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}}}$$
and, of course, x= a- b.