Cubic equation and Cardano's Formula

We know the quadratic formula for the solution of degree two equations

ax² + bx + c = 0

with a,b,c in a field containing rationals. A formula for the solution of such equations expresses the roots of the equations as functions of the coefficients through square root extraction operations

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

We now focus on the resolution of cubic equations with complex coefficients of the form

x³ + ax² + bx + c = 0 (5.8.1)

Then we shall study a formula for solving cubic equations with real coefficients.

The x² term, in 5.8.1 can be eliminated via the substitution x = y − a/3, which results in:

(y − a/3)³ + a(y − a/3)² + b(y − a/3) + c = y³ + (b − a²/3)y + c − ab/3 + 2a³/27 = 0

thus solving 5.8.1 is equivalent to solving the following

y³ + py + q = 0 (5.8.2)

where

p = b − a²/3, q = c − ab/3 + 2a³/27

called the reduced cubic equation or depressed cubic, in which the squared term is missing, a reduction due to Cardano and Ferrari. We will solve it for any coefficient p and q.

The first step towards solving 5.8.2 involves noticing that

(u + v)³ = u³ + 3u²v + 3uv² + v³ = 3uv(u + v) + (u³ + v³)

(u + v)³ − 3uv(u + v) − v³ − u³ = 0

which compared to (5.8.2), pattern matching suggests that if we are able to find two numbers u and v such that

−3uv = p, −u³ − v³ = q (5.8.3)

then the number u + v is a solution of 5.8.2. We can rewrite these two conditions as

u³v³ = −p³/27, u³ + v³ = −q

Since the previous equations are respectively the product and sum of u³ and v³, there is a quadratic equation with roots u³ and v³. Indeed consider the quadratic equation

(x − u³) (x − v³) = x² − xv³ − xu³ − v³u³ = x² − x(u³ + v³) + v³u³

this equation has clearly the solutions v³ and u³. We can write it as

x² + qx −p³/27 = 0

whose roots are

−q/2 ± (q²/4 + p³/27)^1/2

thus

u³ = −q/2 + (q²/4 + p³/27)^1/2 (5.8.4)

and

v³ = −q/2 − (q²/4 + p³/27)^1/2 (5.8.5)

Let u₁, u₂, u₃ and v₁, v₂, v₃ the roots of 5.8.4. and 5.8.5 respectively. We have

u_{1} = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}, u_{2} = ω u_{1}, u_{3} = ω^{2} u_{1}

v_{1} = \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}, v_{2} = ω v_{1}, v_{3} = ω^{2} v_{1}

with ω primitive cubic root of unity, ω = −1/2 +√3i/2, ω² = −1/2 − √3i/2.

The roots of y³ + py + q = 0 are thus given by the following formula, known as Cardano's formula published by Cardano in his Ars Magna of 1545:

y = u + v = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}

where u and v are two solutions of 5.8.4. and 5.8.5. At first glance it may appear that since u and v can assume each three values, u + v could assume more than three values; Out of the nine combinations of these values of u and v, only three cases of u + v will satisfy 5.8.3.

Let u₁ one of the three possible values of u and let v₁ the corresponding value such that u₁v₁ = −p/3. Then the value of v corresponding to u₂ is v₃, because

u₂v₃ = u₁ω ⋅ v₁ω² = u₁v₂ω³ = u₁v₁ = −p/3

whilst for any other choice we do not obtain the desired result

u₂v₂ = u₁ωv₁ω = u₁v₁ω² = −pω²/3 ≠ −p/3

Analogously the value of v corresponding to u₃ is v₂. The three roots of the depressed equation are thus

y₁ = u₁ + v₁, y₂ = u₁ω + v₁ω², y₃ = u₁ω² + v₁ω

Cubic equations with real coefficients

We now want to verify whether Cardano's formula, in the case of a cubic equation with real coefficients gives information on the nature of its roots. A major role is played by the sign of the expression

q²/4 + p³/27

which appears under the square root. The following expression is known as the discriminant of the depressed cubic equation x² px + q = 0 is

Δ = −4p³ − 27q² = −108(q²/4 + p³/27)

it has opposite sign to q²/4 + p³/27.

Let's examine the three cases that can be found using Cardano's formula.

Δ < 0: In this case we have a positive sign under the square root, we thus have to extract the cube root of a real number that we know lead to a real solution and two complex conjugate values. Indicating with u₁ the real value the corresponding value v₁ has to be real as well to have u₁v₁ = −p/3 ∈ ℝ. Hence u₁ + v₁ is real. The other two roots are

y₂ = u₁ω + v₁ω² = u₁ (−1/2 + i√3/2) + v₁ (−1/2 − i√3/2) = −(u₁ + v₁)/2 + i√3 (u₁ − v₁)/2

y₃ = u₁ω² + v₁ω = (u₁ + v₁)/2 − i√3 (u₁ − v₁)/2

they are complex conjugate (and not real).
Δ = 0, In this case

u = (−q/2)^1/3, v = (−q/2)^1/3

If u₁ is the real value of the cubic root, the corresponding v₁ value is real as well, since u₁v₁ = −p/3 and u₁ = v₁. The roots are

y₁ = u₁ + v₁ = 2u₁, y₂ = u₁(ω + ω²) = −u₁, y₃ = u₁(ω² + ω) = −u₁.

All the roots are real and two are equal.
Δ < 0. In this case the expression under the square root is negative, so we extract the cubic root of a complex number, which are complex as well. Thus all values u + v are be complex. A cubic equation always has at least one real root. Suppose this real root to be y₁ = u₁ + v₁. Then u₁ and v₁ are complex conjugate, since both their product and sum are real numbers. Similarly we have also the complex conjugates u₁ω with v₁ω², and u₁ω² with v₁ω. Thus y₁ = u₁ + v₁, y₂ = u₁ω + u₁ω² and y₃ = u₁ω² + v₁ω are all real and distinct: indeed if we take y₂ = y₃ we would have u₁(ω − ω²) = v₁(ω − ω²) implying u₁ = v₁ which is an absurd. Hence in this case all roots are real and distinct.

Focusing one this last case of a cubic equations with all real roots, Cardano's formula led us to the extraction of cubic roots of complex numbers that we are able to find thanks to the trigonometric representation of complex numbers. For this reason the practical applicability of Cardano's formula is limited. This case has historically been known as the “casus irreducibilis”, because of the difficulty in showing that these immaginary terms reduce to real terms.

Example 5.8.1. If we take the depressed cubic equation x³ = 7x + 6, and use the standard recipe for its solution, we obtain a negative number under the sign of the square root i.e. the casus irreducibilis. It is easy to find out the roots −1, −2 and 3. Cardano's formula with p = −7 and q = −6 give us the roots

[3 + (36/4 − 343/27)^1/2)]^1/3 + [3 − (36/4 − 343/27)^1/2)]^1/3 = [3 + (−100/27)^1/2)]^1/3 + [3 − (−100/27)^1/2)]^1/3

To sum up

The roots of y³ + py + q = 0, p,q ∈ ℝ with Δ = −4p³ − 27q²

Δ < 0 one real root and two non-real complex conjugate roots.
Δ = 0 all real roots with two equal.
Δ > 0 the cubic has three distinct real roots

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