The Cubic Equation Formula
WARNING: Do not assign this in MAT 150 for a grade.
The Cubic Equation
f ( x )  x  bx  cx  d  0
3
2
Let:
b

g( y)  f  y  
3


 2b

b 
bc
3
3
 y  c 
y



d

y
 py  q  0



3 
3

 27

2
3
We wish to solve the reduced equation:
h ( y )  y  py  q  0 .
3
Next let:
y  z
p
.
3z
Then:
3
3
p 
p 
p


3
h(z)   z 
q0
  pz 
q  z 
3
3z 
3z 
27 z


Clearing the fractions gives us:
z  qz 
6
3
p
3
27
0
By the Quadratic Formula:
q 
q 
2
4p
3
27
z 
3
 
2
Let  
p
3

q
27
q

2
.
4
Then the two values of z3 are:

2

,

q
2

3
27
2
q
p


q
2
4
Then:
 q
 
 2
3

q
  q
   
  2


3

2
q
  p
    
  3

  
2
3
If we let:
z 
3

q


2
then:

z

3

q
2

p

p
   
3

3
Since:

z

3

q

2

p
   
3

we get:
3

q

  
2
p
.
3z
Since y = z + (p/3z), we get the solution:
y 
3

q
2

 
3

q
2

 .
From
 
p
3

27
q
2
y 
and
3

q

 
3

2
4
q

 ,
2
the real (since p, q are real) solution to the reduced equation
y3 + py + q = 0
is:
y 
3

q
2

p
3
27

q
2
4

3

q
2

p
3
27

q
2
4
.
Relabel:
z 
3

q

  A;
3

2
q

  B
2
So if y3 + py + q = 0, we can rewrite the real solution as:
y 
3

q
2

p
3
27

q
2

3
4
But there should be two others!

q
2

p
3
27

q
2
4
 A B .
By DeMoivre’s Theorem, if A is the real cube root of:

q

2
p
3

27
q
2
,
4
then  A and  2 A are the other two (complex) roots, where:
 
1
3
i
2
2 i
e
3
.
2
Similarly, since B is the real cube root of

q
2

p
3
27

q
2
,
4
then the other two (complex) roots are:  B and  2 B .
.
The result
y=A+B
as a solution of:
y3 + py + q = 0
will also hold if A and B are replaced by the other respective
cube roots, so long as the product of the terms is AB = p/3.
Since 3 = 1, this will also hold true for the pairs
 A,  B
2
and
 A,  B .
2
We get that the three solutions of:
y3 + py + q = 0
are:
y  A  B,  A   B,
2
 A  B
2
where:
A
3

q
2

p
3
27

q
2
and
B 
4
3

q

2
Add b/3 to each solution to get the solutions of:
x3 +bx2 + cx + d = 0.
p
3
27

q
2
4
Example: x3  3x + 2 = 0
The equation is in reduced form with p = 3 and q = 2.
x 

3
3


q
p

3

2
27
2
(  3)

q
3

27
1  1 
3
1 

3
1 
3

3

q
p

2
2

3

3
1 

1  1
1  1  1  2
So x1 = 2 is our first solution.
2
2
4
3
27
2
4
2

2

q
2
4
(  3)
27
3

2
2
4
Using our previous notation, x1 = A + B, where A = B = 1.
The second solution is x2 = A + 2B, where
2 i
 e
Thus:
3
 
1
2
 1
3
x2     i
 (  1) 
2 
 2
i
3
.
2
 1
3
 i
 (  1)  1.
2 
 2
Similarly the third solution x3 = 2A + B = 1.
Thus the three solutions of x3  3x + 2 = 0 are: x = 2, 1, 1.
Example x3 +12x2 + 54x + 68 = 0
Here we have b = 12, c = 54, and c = 68.
We transform to reduced form by letting x = y  b/3 = y  4.
Reduced Form: y3 + 6y  20 = 0. (Cardano’s Example)
Here p = 6 and q = 20.
x1 
x1 
3

q
2
3

p
3

27
10  6 3 
q
2

3

4
3
10  6 3
q
2

p
3
27

q
2
4
Can
3
10  6 3 
3
10  6 3
be made to make sense??
YES!!
Let us start by assuming that:
3
10  6 3  a  b 3 ,
Cubing both sides yields:
10  6 3  ( a  9 ab )  (3 a b  3 b ) 3 .
3
2
2
Equating coefficients and factoring yields:
a (a  9b )  10
2
2
b(a  b )  2
2
2
This works for a = b = 1. Thus:
3
10  6 3  1 
3.
3
Similarly:
3
10  6 3  1 
3.
So:
y1 
3
10  6 3 
3

10  6 3  1 
 
3  1
Letting:
A 1
3,
we see that y1 = A + B, as before.
B 1
3,

3  2
Now, for the other solutions:
y2   A   B
2
 1
3
   i
 1
2 
 2

 1
3
3   i
 1
2 
 2
3   1  3i
 1
3
3    i
 1
2 
 2
3   1  3i



and:
y3   A   B
2
 1
3
  i
 1
2 
 2




Thus the three solutions of
y3 + 6y  20 = 0
are:
y = 2, 1  3i.
Remember that x = y  4. Thus the solutions to the original
equation:
x3 +12x2 + 54x + 68 = 0
are:
x   2,  5  3i
Chronology of the Cubic
1494
In his book Summa de Arithmetica, the Italian Luca
Pacioli states that, given the state of mathematics, the
general cubic ax3 + bx2 + cx + d = 0 is unsolvable.
c. 1500
Scipione del Ferro solves x3 + px = q, x3 + q = px and
x3 = px + q, where p, q > 0. He does not publish it.
c. 1535
Del Ferro reveal his result to his colleague Antonio
Maria Fior.
1535
Fior challenges the Venetian mathematician Niccolò
Fontana (aka “Tartaglia”) to a contest where each poses
thirty math problems to the other. The loser is to pay
for a dinner for thirty. All of Fior’s problems involve
the above cubics.
Chronology of the Cubic
1535 On the night of February 1213, Tartaglia solves all
thirty problems. He generously passes on the dinner.
1539 The scholar Girolamo Cardano asks Tartaglia (through a
third party) what his solution to the cubic is. Tartaglia
refuses. Cardano then invites Tartaglia to Milan as his
guest. He offers Tartaglia the opportunity to show off his
military inventions to the commander of Milan. (He also
gives Tartaglia a dinner.)
1539
On March 29, Tartaglia reveals his solution to x3 + px = q
to Cardano. He swears Cardano to an oath of secrecy.
Cardano extends the result to general cubic equations.
Chronology of the Cubic
1545
Cardano publishes Tartaglia ’s result, along with that of his
protégé, Ludovico Ferrari. Ferrari shows how to use the
reduced cubic to solve quartic equations. Cardano cites
del Ferro and Tartaglia in his treatise.
1548 A furious Tartaglia sues. Ferrari issues a public challenge.
The case is argued in a Milan court on August 10th.
Tartaglia leaves before the case is settled. He loses.
2014
A group of scholars at a KY teaching conference learns
about Cardano’s (and Tartaglia’s, and del Ferro’s) formula.
Bibilography
Dobbs, David and Hanks, Robert. A Modern Course on the Theory of
Equations. Polygonal Publishing House, 1980.
Dunham, William. Journey Through Genius: The Great Theorems of
Mathematics. Wiley, 1990.
Irving, Ron. Beyond the Quadratic Formula. MAA Inc., 2013