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Deriving a Formula for Cubic Roots

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Date: 05/22/98 at 08:44:01
From: Chris Renton
Subject: cubic formula

Hi there,

I was just wondering if there is a working formula to solve for the 3
roots of a cubic. I have tried modifying the "completing the square"
to a "completing the cube" but I end up with "x" on both sides of the
equation. My work so far is below:

I started out by using numbers instead of variables, so I could see
where the pieces of the equation were going. I started with the
perfect cube of (x + 10)^3 = 0, and expanded to:

x^3 + 30x^2 + 300x + 1000 = 0

(I am using the general form of ax^3 + bx^2 + cx + d = 0). Then I
realised that by using a method similar to completing the square, you
can do away with the last term (in this case, last 2 terms). So I
started working backwards from x^3 + 30x^2 + 300x + 1000 = 0.

Completeing the cube:

x^3 + 30x^2 = -300x - 1000

I got the rest of the Left Hand Side by:

x^3 + 30x^2 + ((b/3)^2)x + (b/3)^3 =
-300x - 1000 - ((b/3)^2)x) - ((b/3)^3)

(b = 30, from the co-efficient of x^2)

Then:

x^3 + 30x^2 + 300x + 1000 = -300x - 1000 + 300x + 1000

x^3 + 30x^2 + 300x + 1000 = 0

Now, that it is a perfect cube, so I can factorise:

(x + (b/3))^3 = 0

(x + 10)^3 = 0

x = -10

And the root(s) are found!

The problem arises when the equation isn't a perfect cube, such as:

x^3 + 27x^2 + 243x + 729 = 0

x^3 + 27x^2 = -243x - 729

Following the above method:

x^3 + 27x^2 + ((b/3)^2)x + ( b/3)^3 =
-243x - 729 + ((b/3)^2)x + ((b/3)^3)

x^3 + 27x^2 + 252x + 738 = -243x - 729 + 252x + 738

x^3 + 27x^2 + 252x + 738 = 9x - 9

(x + (b/3))^3 = 9x-9

(x + 9)^3 = 9x - 9

x + 9 = (9x - 9)^(1/3)

See? There is an x on both sides.

So if you've had the patience to follow this so far, I would be very
grateful for any insight on this problem.

ax^3 + bx^2 + cx + d = 0:

Dividing through by the co-efficient of x^3:

x^3 + (b/a)x^2 + (c/a)x + (d/a) = 0/a

x^3 + (b/a)x^2 = -(c/a)x - (d/a)

x^3 + (b/a)x^2 = (-cx - d)/a

x^3 + (b/a)x^2 + ((b/3a)^2)x + (b/3a)^3 =
((-cx - d)/a) + ((b/3a)^2)x + (b/3a)^3

which then expands and simplifies to:

(x + (b/3a))^3 = (-27cxa^2 - 27da^2 + 3xab^2 + b^3)/27a^3

Taking the cube root:

x + (b/3a) = ((-27cxa^2 - 27da^2 + 3xab^2 + b^3)/27a^3)^(1/3)

x + (b/3a) = (-27cxa^2 - 27da^2 + 3xab^2 + b^3)^(1/3)/3a

x = (-b + (-27cxa^2 - 27da^2 + 3xab^2 + b^3)^(1/3))/3a

This is the formula, but x is on the other side (twice). Also this
looks like it will only find one root because there is no
+/- with cube root as there is with square root.

If you've understood this I congratulate you because my maths teacher
at school can't help me.

Very kind regards,
Chris Renton
```

```
Date: 05/22/98 at 09:22:50
From: Doctor Rob
Subject: Re: cubic formula

This is very interesting, but the appearance of x on both sides of the
equation is fatal. You started down the correct path, however.

For one solution technique, see:

http://mathforum.org/dr.math/faq/faq.cubic.equations.html

For another, proceed as in the above page until you have the equation
in the form:

y^3 + p*y + q = 0

This is done as you have shown above by completing the cube, that is,
by setting x = y - b/(3*a), substituting, expanding, and dividing the
equation by a. The next step is to let y = z - p/(3*z). Substituting,
expanding,and multiplying through by z^3, we get:

z^6 + q*z^3 - p^3/27 = 0

This is a quadratic equation in z^3, which you can solve using the

-q +/- sqrt[q^2 + (4*p^3/27)]
z^3 = -----------------------------
2

Now you have to find the cube roots of these two numbers to determine
z. (Actually, you can pick either of these and ignore the other.) If
you can find one of the cube roots of any number, the other two are
gotten by multiplying it by w = (-1+sqrt[-3])/2 and
w^2 = (-1-sqrt[-3])/2.  You will get six values of z, which, when
substituted in y = z - p/(3*z), will give you only three values of y.
These are the three roots of the equation y^3 + p*y + q = 0. From them
you can calculate the roots x of the original equation:

a*x^3 + b*x^2 + c*x + d = 0

via x = y - b/(3*a).

equation has real coefficients and three real roots, the values of z
are always complex, never real. The hardest part of this procedure is
taking the cube root of a complex number.

-Doctor Rob,  The Math Forum
Check out our web site http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents

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