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.