# Solving Cubic Equations

1. Jun 27, 2008

### mikee

1. The problem statement, all variables and given/known dataI was just wondering if anyone could give me a quick lesson on solving cubic equations(Ax^3+Bx^2+Cx+D=0), or a good place to go to, if anyone can help thanks alot

2. Relevant equations

3. The attempt at a solution

2. Jun 27, 2008

### HallsofIvy

Staff Emeritus
here's the way I would argue it:

First, convert from the general cubic to the "reduced cubic" (without the x2 term). If we replace x by y- u, then
$$Ax^3+ Bx^2+ Cx+ D= A(y-u)^3+ B(y- u)^2+ C(y- u)+ D= Ay^3- 3Auy^2+ 3Au^2y- Au^3+ By^2- 2Buy+ Bu^2+ Cy- Cu+ D$$
$$= Ay^3+ (-3Au+ B)y^2+ (3Au^2- 2Bu+ C)y+ (-Au^3+ Bu^2- Cu)+ D[/itex] Choose u so that the coefficient of y2 is 0: That is, choose u= B/3A so that the equation for y has no y2 term. Solve that equation for y, and then x= y+ u. Now to solve that reduced equation: If a and b are any two numbers, then [tex](a+b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$$
and
$$3ab(a+b)= 3a^2b+ 3ab^2$$
so
$$(a+ b)^3- 3ab(a+b)= a^3+ b^3$$
or, letting x= a+b, m= 3ab, and n=a3+ b3,
$$x^3- mx= n$$

Given any "reduced" equation Ax3+ Bx+ C= 0, We can divide through by A to get x3+ (B/A)x+ C= 0 or x3- (-B/A)x= -C. m= -B/A and n= -C/A.

Given m and n, can we solve for a and b, and so find x?

Of course, we can! From m= 3ab, b= m/3a. Replacing b by that in n= a3+ b3, n= a3+ m3/(33a3).l

Multiply through by a3 to get na3= a6+ m3/3 which we can write as a quadratic equation for a3:
(a3)3- na3+ m3/33 and solve that with the quadratic formula:

$$a^3= \frac{n\pm\sqrt{n^2- 4m^2/3^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2-\left(\frac{m}{3}\right)^3}$$

Of course, once you have found a, b= m/3a and x= a+ b.

Using that is a heck of a lot of work which is why it is not normally taught in basic algebra!

Here's a website where they just give the formula:
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

3. Jun 27, 2008

### mikee

hmm thats a little bit more complicated than i thought but thank you