🧊 Exploring Cubic

Here we explore how to solve a cubic equation x³ + a x² + b x + c = 0, looking for a general solution. 

1. Depressing the cubic

The goal is to reduce a cubic equation to one without the x² term, such that the coefficient of x² is 0.

Starting from x³ + a x² + b x + c = 0,
we want to obtain y³ + py + q = 0.

Let x = y + k

(y + k)³ + a (y + k)² + b (y + k) + c = 0
y³ + 3y²k + 3yk² + k³ + a (y² + 2yk + k²) + b (y + k) + c = 0
y³ + (3k + a) y² + (3k² + 2ak + b) y + (k³ + ak² + bk + c) = 0

To make the coefficient of y² equal to 0:
3k + a = 0
k = - a / 3

Let p = 3k² + 2ak + b
p = 3(- a / 3)² + 2a(- a / 3) + b
p = a²/ 3 - 2a²/ 3 + b
p = b - a²/ 3

Let q = k³ + ak² + bk + c
q = (- a / 3)³ + a(- a / 3)² + b(- a / 3) + c
q = - a³ / 27 + a³ / 9 - ab / 3 + c
q = 2 a³ / 27 - ab / 3 + c

Thus, by substituting:

• x = y - a / 3

• p = b - a²/ 3

• q = 2 a³ / 27 - ab / 3 + c

into x³ + a x² + b x + c = 0, we obtain the depressed cubic y³ + py + q = 0.

4. Connecting the Dots

Back to:
m³ = (- q ± √(q² + 4p³ / 27) ) / 2

For simplicity, call the right-hand side A,
m³ = A

Therefore, m = ∛ A

Since w³ = 1,
w³ * m³ = 1 * A
(wm)³ = A
wm =  ∛ A
So, wm is another cube root of A.

Similarly w⁶ = 1,
w⁶ * m³ = 1 * A
(w²m)³ = A
w²m =  ∛ A
Likewise, w²m is another cube root of A.

Thus, the 3 cube roots of A = { m, wm, w²m }.

Since n = p / 3 m, and y = m - n, root of the depressed cubic are:

• y₁ = m - p / 3 m

• y₂ = wm - p / 3 wm

• y₃ = w²m - p / 3 w²m

Knowing that 1 / w = w² and 1 / w² = w, the roots of the depressed cubic can be written more elegantly as:

• y₁ = m - n

• y₂ = wm - w²n

• y₃ = w²m - wn

5. Resolving the ± Ambiguity 

Back to:
m³ = (- q ± √(q² + 4p³ / 27) ) / 2

The ± is not a choice between the three roots. The other two values of m are then generated using the cube roots of unity. For the remainder of this derivation, we choose the + branch.

Thus,

• m = ∛ ( (- q + √(q² + 4p³ / 27) ) / 2 )

• n = p / (3 ( ∛ ( (- q + √(q² + 4p³ / 27) ) / 2 ) ) )

6. Euler Comes In

What if the discriminant q² + 4p³ / 27 ​< 0 ?

Starting from:
m³ = (- q + √(q² + 4p³ / 27) ) / 2

Let discriminant u = q² + 4p³ / 27,
m³ = (- q + √u) / 2

Since u < 0, √ u = i √ - u:
m³ = (- q + i √ - u) / 2
m³ = - q / 2 + i (√ - u / 2)

Convert the right-hand side to Euler's form such that re^(i Θ), where:

• r = √( (- q / 2)² + (√ - u / 2)²)
  r = √( q² / 4  - u / 4)
  r = √(q² - u) / 2
  r = √(q² - (q² + 4p³ / 27) ) / 2
  r = √(- 4 p³ / 27) / 2
  r = √(- p³ / 27)

• tan Θ = (√ - u / 2) / (- q / 2)
  tan Θ = √ - u / - q

A note that as r > 0, ∛ r = r^(1/3).

m³ = re^(i Θ)
m = ∛ (re^(i Θ))
m = r^(1/3) e^(i Θ / 3)

Applying cube root of unity, the cube root of r e^(i Θ):

• m = r^(1/3) e^(i Θ / 3)

• wm = wr^(1/3) e^(i Θ / 3)

• w²m = w²r^(1/3) e^(i Θ / 3)

Rewriting w and w² in Euler's Form:

• w = - 1 / 2 + i √ 3 / 2
  w = e^(i 2π / 3)

• w² = - 1 / 2 - i √ 3 / 2
  w² = e^(i 4π / 3)

Thus, the cube root of r e^(i Θ):

• m = r^(1/3) e^(i Θ / 3)

• wm = e^(i 2π / 3) * r^(1/3) e^(i Θ / 3)
  wm = r^(1/3) e^(i (2π + Θ) / 3)

• w²m = e^(i 4π / 3) * r^(1/3) e^(i Θ / 3)
  w²m =  r^(1/3) e^(i (4π + Θ) / 3)

As for n = p / 3m:

• n = p / (3 r^(1/3) e^(i Θ / 3) )
  n = p / 3 * r^(-1/3) e^(- i Θ / 3)

• w²n = e^(i 4π / 3) * p / 3 * r^(-1/3) e^(- i Θ / 3)
  w²n = p / 3 * r^(-1/3) e^(i (4π - Θ) / 3)

• wn = e^(i 2π / 3) * p / 3 * r^(-1/3) e^(- i Θ / 3)
  wn =  p / 3 * r^(-1/3) e^(i (2π - Θ) / 3)

Roots of the depressed cubic:

• y₁ = m - n
  y₁ = r^(1/3) e^(i Θ / 3) - p / 3 * r^(-1/3) e^(- i Θ / 3)

• y₂ = wm - w²n
  y₂ = r^(1/3) e^(i (2π + Θ) / 3) - p / 3 * r^(-1/3) e^(i (4π - Θ) / 3)

• y₃ = w²m - wn
  y₃ = r^(1/3) e^(i (4π + Θ) / 3) - p / 3 * r^(-1/3) e^(i (2π - Θ) / 3)

Since r = √(- p³ / 27):
r = (- p³ / 27) ^ (1/2)
r^(1/3) = ((- p³ / 27) ^ (1/2)) ^ (1/3)
r^(1/3) = ((- p³ / 27) ^ (1/3)) ^ (1/2)
r^(1/3) = (- p / 3 ) ^ (1/2)
r^(1/3) = √ (- p / 3 )

(- p / 3) = r^(2/3)
(- p / 3) * r^(-1/3) = r^(2/3) * r^(-1/3)
- p / 3 * r^(-1/3) = r^(1/3)

The first root of the depressed cubic y₁:
y₁ = r^(1/3) e^(i Θ / 3) - p / 3 * r^(-1/3) e^(- i Θ / 3)
y₁ = r^(1/3) e^(i Θ / 3) + r^(1/3) e^(- i Θ / 3)
y₁ = r^(1/3) (e^(i Θ / 3) + e^(- i Θ / 3) )

Knowing the identity e^(i Θ) + e^(- i Θ) = 2 cos Θ, then:
e^(i Θ / 3) + e^(- i Θ / 3) = 2 cos (Θ / 3)

y₁ = r^(1/3) * 2 cos (Θ / 3)
y₁ = 2 r^(1/3) cos (Θ / 3)
The imaginary parts cancelled out leaving a real root for y₁.

What about y₂ and y₃?

• y₂ = r^(1/3) e^(i (2π + Θ) / 3) - p / 3 * r^(-1/3) e^(i (4π - Θ) / 3)
  y₂ = r^(1/3) e^(i (2π + Θ) / 3) + r^(1/3) e^(i (4π - Θ) / 3)
  y₂ = r^(1/3) * (e^(i (2π + Θ) / 3) + e^(i (4π - Θ) / 3))

• y₃ = r^(1/3) e^(i (4π + Θ) / 3) - p / 3 * r^(-1/3) e^(i (2π - Θ) / 3)
  y₃ = r^(1/3) e^(i (4π + Θ) / 3) + r^(1/3) e^(i (2π - Θ) / 3)
  y₃ = r^(1/3) * (e^(i (4π + Θ) / 3) + e^(i (2π - Θ) / 3))

Casus irreducibilis payoff:

• y₁ = 2 r^(1/3) cos (Θ / 3)

• y₂ = -2 r^(1/3) cos ((Θ - π) / 3)

• y₃ = -2 r^(1/3) cos ((π + Θ) / 3)

8. Determining the Nature of Roots

The discriminant u = q² + 4p³ / 27. When:

• u > 0, 1 real + 2 complex conjugates. The other two roots are obtained using the cube roots of unity. Ugly complex number ahead!

• u = 0, All real. At least 2 are identical. Easiest by far!

• u < 0, 3 distinct real (casus irreducibilis). Apparently all real roots ended up with the most complex of calculation. What a paradox. 🥴

9.1 Example 1 - Cubic with 1 Real Root + 2 Complex Conjugate Roots

Let's build a cubic with roots x = { 2, 3 - 4i, 3 + 4i }.

(x - 2) (x - (3 - 4i)) (x - (3 + 4i)) = 0
(x - 2) (x²- (3 - 4i) x - (3 + 4i) x + (3 - 4i) (3 + 4i) ) = 0
(x - 2) (x² - 6 x + 9 + 16) = 0
(x - 2) (x² - 6 x + 25) = 0
x³ - 8x² + 37x - 50 = 0

x³ + a x² + b x + c = 0
a = - 8
b = 37
c = - 50

Depressing the cubic:

• x = y - a / 3
  x = y - (- 8) / 3
  x = y + 8 / 3

• p = b - a² / 3
  p = 37 - (- 8)² / 3
  p = 37 - 64 / 3
  p = 47 / 3

• q = 2 a³ / 27 - ab / 3 + c
  q = 2 (- 8)³ / 27 - (- 8) (37) / 3 + (- 50)
  q = - 2 (512) / 27 + 296 / 3 - 50
  q = - 1024 / 27 + 2664 / 27 - 1350 / 27
  q = 290 / 27

Thus, the depressed cubic:
y³ + (47 / 3) y + 290 / 27 = 0 

Running the discriminant u = q² + 4p³ / 27:
Since q² > 0, p > 0, thus u > 0. This is the 1 real + 2 complex conjugates case.

Calculate u = q² + 4p³ / 27:
u = (290 / 27)² + 4 (47 / 3)³ / 27
u = 18496 / 27

From m³ = (- q + √u) / 2:
m³ = (- 290 / 27 + √ (18496 / 27) ) / 2
m ≈ 1.976067743

Then, n = p / 3 m:
n ​= (47 / 3) / 3 m
n ​= 47 / 9 m
n ​≈ 47 / (9 * 1.976067743)
n ≈ 2.642734411

Note that cube roots of unity:

• w = - 1 / 2 + i √ 3 / 2

• w² = - 1 / 2 - i √ 3 / 2

Calculating y:

• y₁ = m - n
  y₁ ≈ 1.976067743 - 2.642734411
  y₁ ≈ - 0.66667

• y₂ = wm - w²n
  y₂ ≈ (- 1 / 2 + i √ 3 / 2) * 1.976067743 - (- 1 / 2 - i √ 3 / 2) * 2.642734411
  y₂ ≈ 0.33333334 + 4i

• y₃ = w²m - wn
  y₃ ≈ (- 1 / 2 - i √ 3 / 2) * 1.976067743 - (- 1 / 2 + i √ 3 / 2) * 2.642734411
  y₃ ≈ 0.333333334 - 4i

Back to x = y + 8 / 3:

• x₁ ≈ - 0.66667 + 8 / 3
  x₁ ≈ 2

• x₂ ≈ 0.33333334 + 4i + 8 / 3
  x₂ ≈ 3 + 4i

• x₃ ≈ 0.33333334 - 4i + 8 / 3
  x₃ ≈ 3 - 4i

x = { 2, 3 + 4i, 3 - 4i }

9.2 Example 2 - Cubic with All Real Roots & at least 2 Identical

Starting with roots x = { 2, 3, 3 }.

(x - 2) (x - 3) (x - 3) = 0
(x - 2) (x² - 6 x + 9) = 0
x³ - 8 x² + 21 x - 18 = 0

x³ + a x² + b x + c = 0
a = - 8
b = 21
c = - 18

Depressing the cubic:

• x = y - a / 3
  x = y - (- 8) / 3
  x = y + 8 / 3

• p = b - a² / 3
  p = 21 - (- 8)² / 3
  p = 21 - 64 / 3
  p = - 1 / 3

• q = 2 a³ / 27 - ab / 3 + c
  q = 2 (- 8)³ / 27 - (- 8) (21) / 3 + (- 18)
  q = - 2 (512) / 27 + 168 / 3 - 18
  q = 2 / 27

Hence, the depressed cubic:
y³ - (1 / 3) y + 2 / 27 = 0

Running the discriminant u = q² + 4p³ / 27:
u = (2 / 27)² + 4 (- 1 / 3)³ / 27
u = 0, leading to all 3 real roots with at least 2 identical roots.

From m³ = (- q + √u) / 2:
m³ = - q / 2
m³ = - q / 2
m³ = - (2 / 27) / 2
m³ = - 1 / 27
m = - 1 / 3

Then, n = p / 3 m:
n = (- 1 / 3) / 3 (- 1 / 3)
n = 1 / 3

Note that cube roots of unity:

• w = - 1 / 2 + i √ 3 / 2

• w² = - 1 / 2 - i √ 3 / 2

Calculating y:

• y₁ = m - n
  y₁ = - 1 / 3 - 1 / 3
  y₁ = - 2 / 3

• y₂ = wm - w²n
  y₂ = (- 1 / 2 + i √ 3 / 2) * (- 1 / 3) - (- 1 / 2 - i √ 3 / 2) * (1 / 3)
  y₂ = 1 / 6 + 1 / 6 - (1 / 3) (i √ 3 / 2) + (1 / 3) (i √ 3 / 2)
  y₂ = 1 / 3

• y₃ = w²m - wn
  y₃ = (- 1 / 2 - i √ 3 / 2) (- 1 / 3) - (- 1 / 2 + i √ 3 / 2) * (1 / 3)
  y₃ = 1 / 6 + 1/ 6 + (1 / 3) (i √ 3 / 2) - (1 / 3) (i √ 3 / 2)
  y₃ = 1 / 3

The complex part of y₂ and y₃ beautifully conjugated and cancelled out each other, leaving only the same real part.

Calculating x = y + 8 / 3 :

• x₁ = - 2 / 3 + 8 / 3
  x₁ = 2

• x₂ = 1 / 3 + 8 / 3
  x₂ = 3

• x₃ = 1 / 3 + 8 / 3
  x₃ = 3

x = { 2, 3, 3 }

9.3 Example 3 - Cubic with 3 Distinct Real Roots

Let's generate one cubic equation with roots { 2, 3, 6 }.
(x - 2) (x - 3) (x - 6) = 0
(x - 2) (x² -9x + 18) = 0
x³ - 11x²+ 36x - 36 = 0

x³ + a x² + b x + c = 0
a = - 11
b = 36
c = - 36

Depressing the cubic:

• x = y - a / 3
  x = y + 11 / 3

• p = b - a² / 3
  p = 36 - (- 11)² / 3
  p = 36 - 121 / 3
  p = - 13 / 3

• q = 2 a³ / 27 - ab / 3 + c
  q = 2 (- 11)³ / 27 - (- 11)(36) / 3 + (- 36)
  q = - 2662 / 27 + 132 - 36
  q = - 2662 / 27 + 132 - 36
  q = - 2662 / 27 + 96
  q = - 70 / 27

Thus, the depressed cubic:
y³ - (13 / 3) y - 70 / 27 = 0

Running the discriminant u = q² + 4p³ / 27:
u = (- 70 / 27)² + 4 (- 13 / 3)³ / 27
u = 4900 / 729 - 8788 / 729
u = - 3888 / 729

u < 0, this one would land in the casus irreducibilis case, leading to 3 distinct real roots.

From our previous derivation,
r^(1/3) = √ (- p / 3 )
r^(1/3) = √ (-  (- 13 / 3) / 3)
r^(1/3) = √ (13 / 9)
r^(1/3) = √ 13 / 3

From m³ = - q / 2 + i (√ - u / 2):
m³ = - (- 70 / 27) / 2 + i (√ - (- 3888 / 729) / 2)
m³ = 70 / 27 / 2 + i (√ (3888 / 729) / 2)
Both the real and imaginary parts > 0, so m³ lies in Quadrant I of the Argand diagram.

tan Θ = √ - u / - q
tan Θ = √ - (- 3888 / 729) / - (- 70 / 27)
tan Θ = √ 3888 / 70
tan Θ = 36 √ 3 / 70
tan Θ = 18 √ 3 / 35
Θ ≈ 0.7287 rad

y₁ = 2 r^(1/3) cos (Θ / 3)
y₁ = 2 √ 13 / 3 cos (Θ / 3)
y₁ ≈ 2 √ 13 / 3 cos (0.7287 / 3)
y₁ ≈ 2 √ 13 / 3 cos (0.7287 / 3)
y₁ ≈ 2.3333

y₂ = -2 r^(1/3) cos ((Θ - π) / 3)
y₂ = -2 √ 13 / 3 cos ((Θ - π) / 3)
y₂ ≈ -2 √ 13 / 3 cos ((0.7287 - π) / 3)
y₂ ≈ -1.6667

y₃ = -2 r^(1/3) cos ((π + Θ) / 3)
y₃ = -2  √ 13 / 3  cos ((π + Θ) / 3)
y₃ ≈ -2  √ 13 / 3  cos ((π + 0.7287) / 3)
y₃ ≈ -0.6667

Back to x = y + 11 / 3:

• x₁ ≈ 2.3333 + 11 / 3
  x₁ ≈ 7 / 3 + 11 / 3
  x₁ ≈ 6

• x₂ ≈ -1.6667 + 11 / 3
  x₂ ≈ - 5 / 3  + 11 / 3
  x₂ ≈ 2

• x₃ ≈ -0.6667 + 11 / 3
  x₃ ≈ - 2 / 3 + 11 / 3
  x₃ ≈ 3

x = { 2, 3, 6 }

We started with a cubic whose roots were deliberately chosen to be 2, 3, 6. After depressing the cubic and passing through the casus irreducibilis machinery, Cardano's formula brings us back to those same three real roots.

Congratz! That's all to it here. Thanks for sticking out until the end. 🎉