April 17, 2015

Homework Help: Factoring

Posted by Ace on Wednesday, May 30, 2007 at 12:47am.

What does 3x^3-3x+2=0 come out to? How would I solve this?

Is the x cubed supposed to be x squared?

You can compare the equation with the identity:

(a+b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3

which you can rewrite as:

(a + b)^3 = 3 ab (a + b) + a^3 + b^3 (1)

The equation you want to solve is:

x^3 = x - 2 (2)

Equation (1) is an identity that is always true. I.e., whatever you substitute for a and b, the equation is always satisfied. Equation (2) has, of course, at most three solutions. Suppose you choose a and b such that:

3 ab = 1 (3)


a^3 + b^3 = -2 (4)

then equation (1) says that:

(a + b)^3 = (a + b) - 2

and you see that a + b would then satisfy equation (2). So, how do we solve equations (3) and (4) for a and b? You take the cube of equation (3):

3 ab = 1 --->

a^3 b^3 = 1/27

If you put A = a^3 and B = b^3, then:

A B = 1/27 (5)

equation (4) can be written as:

A + B = -2 (6)

If you use (6) to express B in terms of A and substitute that in (5) yo obtain a quadratic equation for A. Solve that equation. You obtain two solutions. Because of the symmetry between A and B (i.e. if you interchange A and B in (5) and (6) the two equatons remain the same), you can take A to be one of the solutions and B to be the other solution.

Extract the cube roots to obtain a and b, add them up to obtain the solution. To obtain the two other solutions, you must use that for given A there are three solutions for a, such that

a^3 = A:

a = A^(1/3) Exp[2 n pi i/3]

where n is 0, 1 or 2. The same is true for b, but we need to satisfy
equation (3). This means that if we choose a to be:

a = A^(1/3) Exp[2 n pi i/3]

then b is given by:

b = B^(1/3) Exp[-2 n pi i/3]

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