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So, You Want to Solve the Cubic?
Dana P. Williams
9 April 2009
Dana P. Williams
So, You Want to Solve the Cubic?
First Simplification
We want to solve the equation
y 3 + by 2 + cy + d = 0.
(1)
This, as we shall see, is not so easy. Therefore, we make a
simplification that has been around for centuries. We let
b
x := y + .
3
Then, solving for x, we get
x 3 − 3px + 2q = 0.
This is the equation we’ll concentrate on now.
Dana P. Williams
So, You Want to Solve the Cubic?
(2)
A little Calculus
Lemma
Consider equation (2) (that is, x 3 − 3px + 2q = 0) with p 6= 0.
(Equation (2) is pretty easy to solve if p = 0.) Then let
D := q2 − p3 .
1
If D > 0, then (2) has exactly one real root.
2
If D = 0, then (2) has exactly two distinct real roots — one
of them is repeated.
3
If D < 0, then (2) has three distinct real roots.
Proof.
This result is proved using a little one variable calculus. I am going
to leave it as a bit of a challenge.
Do it!
Dana P. Williams
So, You Want to Solve the Cubic?
The Next Ancient Trick
To solve (2), we introduce an extra variable:
x = u + v.
Then plugging into (2), we get
(u + v )3 − 3p(u + v ) + 2q = 0.
A little algebra brings us to
u 3 + v 3 + (u + v )(3uv − 3p) + 2q = 0.
(3)
Since we introduced an extra variable, we can impose an additional
requirement. Namely,
uv = p.
Using this condition in (3) gives us the simple equation
u 3 + v 3 = −2q.
Dana P. Williams
So, You Want to Solve the Cubic?
We Need Another Equation Involving u and v
Since u 3 + v 3 = −2q, we can square both sides to get
u 6 + 2u 3 v 3 + v 6 = 4q 2 .
Then we get
(u 3 − v 3 )2 = u 6 − 2u 3 v 3 + v 6
= u 6 + 2u 3 v 3 + v 6 − 4u 3 v 3
which, in view of (4) and uv = p, is
= 4q 2 − 4p 3 = 4D.
1
Thus if D 2 is some choice of square root for D, then
1
u 3 − v 3 = 2D 2 .
Dana P. Williams
So, You Want to Solve the Cubic?
(4)
Solving for u and v
Now we have two equations:
(
u3 + v 3
u3 − v 3
= −2q
1
= 2D 2 .
(5)
Now some easy algebra gives
1
u 3 = −q + D 2
1
and v 3 = −q − D 2 .
1
(6)
Note that a different choice for D 2 just interchanges the formulas
for u 3 and v 3 . Therefore u + v , uv and u 3 + v 3 + 2q don’t care
which choice we make. This just means that we are free to choose
For example, when D > 0, the natural
something convenient.
√
choice is simply D.
Dana P. Williams
So, You Want to Solve the Cubic?
The Case D > 0
Theorem
If D > 0, then the unique real root of (2) is
q
q
√
√
3
3
x1 = −q + D + −q − D.
The remaining (complex roots) can be found
the quadratic
q using √
3
formula (after division), or by letting u1 = −q + D and
q
√
3
v1 = −q − D. Then the complex roots are
√
u1 + v1
3
x2 = −
+i
(u1 − v1 ) and
2
2
Dana P. Williams
x3 = x̄2 .
So, You Want to Solve the Cubic?
The Proof of the First Theorem
Proof.
Let ω := − 21 + i
√
√
3
2 .
Thus the cube roots of 1 are 1, ω and
=
− i 23 . Now
solutions: x = uk + vj
ω2
− 12
uk := ω k−1
taking cube roots in (6) gives 9 possible
(for k, j = 1, 2, 3) where
q
q
√ √ 3
j−1 3
−q + D
and vj := ω
−q − D .
But these give solutions to (2) only when
p
p = uk vj = ω k+j−2 3 q 2 − D = ω k+j−2 p.
Thus the solutions are x1 = u1 + v1 , x2 = u2 + v3 and
x3 = u3 + v2 . This clearly gives the formula claimed for x1 . I’ll
leave the algebra for x2 and x3 to you.
Dana P. Williams
So, You Want to Solve the Cubic?
An Example
Example (Just how great is this?)
Consider the equation x 3 − 3x − 18 = 0. In this example, p = 1 and
q = −9. Since D = q 2 − p 3 = 80 > 0, there is a unique real root.
Namely
q
q
√
√
3
3
x1 = 9 + 80 + 9 − 80.
This is all very exciting until you realize that x1 = 3. For example,
√
√
√
√
(3 + 5)3
72 + 32 5
9 + 80 = 9 + 4 5 =
=
.
8
8
Thus
u1 =
√
q
√
3+ 5
3
9 + 80 =
.
2
Similarly:
v1 =
√
3− 5
.
2
Thus u1 + v1 = 3. You can also check that x2 = − 32 + i
x3 =
− 32
−
√
i 215 .
Dana P. Williams
√
15
2
So, You Want to Solve the Cubic?
and
The Case D = 0
Theorem
If D = 0, then the two distinct real roots of (2) can be computed
√
√
as follows. If q > 0, then the roots are −2 p and p. If q < 0,
√
√
then the roots are 2 p and − p.
Proof.
√
Since D = 0, we have q = ±p p. Now just plug in and check.
Dana P. Williams
So, You Want to Solve the Cubic?
The Cool Case: D < 0
Notice that if D = q 2 − p 3 < 0, then p > 0.
Theorem
Suppose that D < 0. Let
θ := cos−1
−q √ .
p p
Then the three distinct real roots of (2) are given by
θ + 2πk √
xk = 2 p cos
k = 0, 1, 2.
3
Dana P. Williams
So, You Want to Solve the Cubic?
The Proof
√
√
1
We choose D√2 = i −D. Then u 3 = −q + i −D and
v 3 = −q − i −D.
p
p
√
Note that |u|3 = |u 3 | = q 2 + D = p 3 = p p. √
On the other hand, Im(u 3 ) > 0, so θ := cos−1 p−q
p is an
argument for u 3 .
√
Thus the solutions
tou 3 = −q + i −D are given by
√
uk = p exp i θ+2πk
for k = 0, 1, 2.
3
Since Im(v 3 ) < 0, a good choice for the argument of v 3 is −θ,
where θ is the argument
foru 3 .
√
Thus vj = p exp i −θ+2πj
for j = 0, 1, 2.
3
Dana P. Williams
So, You Want to Solve the Cubic?
More of the Proof
Recall that we require that
2π(k + j) p = uk vj = p exp i
.
3
Therefore our solutions should be u0 + v0 , u1 + v2 and u2 + v1 .
Then, for example,
u0 + v0 =
θ
θ √ θ
√
p exp i
+ exp −i
= 2 p cos
.
3
3
3
A bit more work is required to see that u1 + v2 and u2 + v1 given
the formulas in the statement of the theorem.
Dana P. Williams
So, You Want to Solve the Cubic?
An Example with Three Real Roots
Example
Consider x 3 − 9x − 9 = 0. Here p = 3 and q = − 92 . Then
D = 81
0. So we plug into the formula in the theorem to
4 − 27 <√
get θ = cos−1 23 = π6 . So are solutions are
√
π
≈ 3.4115
2 3 cos
18
√
13π 2 3 cos
≈ −2.2267
18
√
25π 2 3 cos
≈ −1.1848.
18
Dana P. Williams
So, You Want to Solve the Cubic?
Just Checking
15
10
5
K3
K2
K1
0
K5
1
2
3
4
x
K10
K15
Figure: The graph of f (x) = x 3 − 9x − 9.
Dana P. Williams
So, You Want to Solve the Cubic?