The Cubic Formula and Derivation

The Cubic Formula and Derivation

Daniel Rui

Here is the general cubic, with the x3 coefficient already divided into the other coefficients, right

hand side already set to zero because we are finding roots: x3 + ax2 + bx + c = 0.

We

substitute

in

x

=

y

-

a 3

to

get

( y3

-

2a y2

+

a2

y

-

a y2

+

2a2

y

-

a3

)

3

93

9 27

+a

( y2

-

2a y

+

a2

)

39

( a) +b y -

3

+c = 0

Simplifying to give

y3-ay2+ a2 y- a3 3 27

+ay2- 2a2 y+ a3 39 ab +by- 3

+c = 0

Simplifying more to give

y3

-

a2 y

+

by

-

a3

+

a3

-

ab

+c

=

0

3

27 9 3

This is of the form y3 + dy + e = 0 where

a2

a3 a3 ab

d = b - and e = - + - + c

3

27 9 3

1

 We substitute y = 3 u - 3 v to get

3 u3

-

3

3 u2

3v

+

3

3u

3 v2

-

3v

3

+

d3u

-

d3v

+

e

=

0

Simplifying to get

(u

-

v)

-

d3v

-

33 u2

3v

+

d3u

+

3

3u

3 v2

+

e

=

0

Anti-distributing to get

(u

-

v)

-

3 v(d

-

33 u2)

+

3 u(d

+

33 v2)

+

e

=

0

Let's

define

v

-

u

=

e

('cause

we

can),

causing

- 3 v(d

-

33 u2)

+

3 u(d

+

33 v2)

=

0,

which

can

be

simplified to

3 u(d

+

33 v2)

=

3 v(d

+

33 u2)

d3u

+

3

3u

3 v2

=

d3v

+

3

3v

3 u2

d3u

-

d3v

=

3

3v

3 u2

-

3

3u

3v

2

d( 3 u - 3 v) = 3 3 v 3 u( 3 u - 3 v)

d=33v3u

d = 3 uv

3 d3

= uv 27

From the first purple equation, we have v = e + u, which we can put into the second to get

u(e + u) = d3 u2 + eu - d3 = 0

27

27

2

Where we can solve for u with the quadratic formula.

-e ? u=

e2

+

4

d3 27

2

We can also get equations for v; the first equation gives u = v - e, which we stuff into the second

equation to get

v(v - e) = d3 v2 - ev - d3

27

27

Quadratic formula yielding

e? v=

e2

+

4

d3 27

2

Via our definition above (y = 3 u - 3 v), we get

3 -e ? y=

e2

+

4

d3 27

-

3

e?

2

e2

+

4

d3 27

2

And

from

the

definition

above

that

(x

=

y

-

a 3

),

we

have

3 -e ? x=

e2

+

4

d3 27

-

3

e?

2

e2

+

4

d3 27

-

a

2

3

However, there is a problem. The ? gives solutions that don't satisfy e = v - u, so we just keep the positive. And finally, if we want, we can plug in d and e to get the cubic formula

x=

3

-(-

a3 27

+

a3 9

-

ab 3

+ c) +

(-

a3 27

+

a3 9

-

ab 3

+ c)2

+

4

(b-

a2 3

27

)3

2

3

-

(-

a3 27

+

a3 9

-

ab 3

+

c)

+

(-

a3 27

+

a3 9

-

ab 3

+

c)2

+

4 (b-

a2 3

)3

27

a -

2

3

This formula only gives one root; using roots of unity we can get the others.

3

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