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The standard method of solving a quadratic equation (ax^2+bx+c=0) is with the Quadratic Formula. A couple of people have asked me if there is a similar method for cubic equations (mz^3+nz^2+oz+p=0). Such an equation has three solutions, at least one of which is real. You probably should try to factor the equation first, as it may have at least one relatively easy solution. For example x^3-2x^2-2x-3=0 has these likely factors: x+1, x-1, x+3, and x-3. Dividing x^3-2x^2-2x-3 by each of them shows that x^3-2x^2-2x-3=(x^2+x+1)(x-3). So 3 is a solution of x^3-2x^2-2x-3=0, and the other two can be found using the quadratic formula. Graphing may also give useful clues toward the solutions.
There are several general, exact methods for solving cubic equations, all of which are a little bit difficult. So a popular method is to actually estimate the three solutions (as accurately as we want) with one of several computer algorithms. Here is one of the exact methods:
Before we begin, we can transform mz^3+nz^2+oz+p=0 into the standard form x^3+ax+b=0 (sometimes given as x^3+ax=b with an opposite sign on b). Divide mz^3+nz^2+oz+p=0 by m to get the equation into the form z^3+fz^2+gz+h=0, then substitute x-f/3 for z to arrive at the form x^3+ax+b=0 [where a=(3f-g^2)/3 and b=(2f^3-9fg+27h)/27]. Once we have solved for x, we can go back and deduce z.
Let A=cuberoot(-b/2+sqr(b^2/4+a^3/27))
and B=cuberoot(-b/2-sqr(b^2/4+a^3/27))Then the three solutions are:
x=A+B, x=-(A+B)/2+(A-B)sqr(-3)/2, and x=-(A+B)/2-(A-B)sqr(-3)/2