Resolving cubic equations by an homography
(or Moebius transformation)
by Sylvain Poirier
French version : Résolution des
équations du 3ème degré
(It seems the best display of this page is by Firefox)
Historical context
I found this solution when I was student in Classe Preparatoire
(intensive undergraduate study in high schools), to apply for the
Prix
Fermat Junior (1995).
This competition is open every 2 years for original works by
French high school and undergraduate students. Sometimes they find
no worthy candidates and don't give the Prize. That time they
found 2 good ones: myself and Y. Ollivier. After hesitation, they
gave the prize to him, but still granted me a Special Mention to
still point out the value of my work, that they found to be
original (not seen elsewhere). I heard they decided so by
consideration of age, as he was younger than I. Then we were all
invited to the Prize ceremony, so were 3 laureates : Andrew Wiles,
getting the big Fermat Prize for his resolution of the Last Fermat
Theorem ; Yann Ollivier, who got the Junior Prize ; and I, who got
the symbolic honor but no money.
It happens that I had made my finding and wrote my article in
about 1 week or the like (I forgot exactly, but less than 2 weeks
anyway) while Y. Ollivier had taken 1 year to make his work.
The present page is only a simplified and shorter version of the
work, compared to what was actually submitted for that Prize
(which started with considerations on trilinear symmetric forms in
ℂ^{2}, where a tensor expression can naturally provide a
bilinear symmetric form out of a trilinear one).
Algebraic method
To resolve a general cubic equation written as
x^{3} + 3ax^{2}
+ 3bx + c = 0 
(1) 
let us write it as
(x − u)^{3}
+ λ(x − v)^{3} = 0 
(2) 
where u ≠ v. It develops as
(1 + λ)x^{3} − 3(u + vλ)x^{2}
+ 3(u^{2} + v^{2}λ)x − (u^{3}
+ v^{3}λ) = 0,
We identify the coefficients of both cubic equations up to a
proportionality factor.
Note that the terms (1 + λ, u + vλ, u^{2}
+ v^{2}λ, u^{3} + v^{3}λ)
are consecutive terms of a recursive sequence with equation
s_{n+2} − (u + v)su_{n+1}
+ (uv)s_{n} = 0
The condition on u and v for the existence of a λ
making both equations proportional, is that the sequence (1, −a,
b, −c) satisfies the same recursive equation
b + (u + v)a
+ uv = 0
c + (u + v)b + uva = 0 
(3)

This linear equations system with unknowns (u + v)
and (uv) is resolved as
u + v = 
c − ab
a²  b

u v = 
b^{2} − ac
a² − b

So u and v are both solutions of the equation
(a^{2} − b)
y^{2} + (ab − c) y + (b^{2}
− ac) = 0 
(4)

Let us first examine both kinds of possible exceptions:
 If (a^{2} − b) = 0 then the equation
(1) is (x+a)^{3} = a^{3} −
c whose solution is immediate.
 If (4) has a double root, then it is also a double root of
(1). Indeed changing the variable with
X = x − 
u + v
2

= x
− 
c − ab
2(a²
−b)

(5)

we see from the form of (2) that the same translation applied to
u and v globally preserves the system (3). We
come to the case u + v = 0. Then the double root
of (4) transported this way becomes u = v = 0,
which reduces (3) to b = c = 0. This result
substituted to (1) gives that 0 is double root there too. As the
same translation was applied and gives the same double root for
both systems, these double roots were thus initially equal too.
Now on these two exceptional cases will be excluded from our study.
Equation (1) will have 3 real roots iff (4) has no real root, for
then we work in the set of complex numbers, and the third root
"function" that shall be used will give the same status to its 3
results; but if the roots of (4) are real then, in ℝ, the third root has a privileged
result (the real one) that will give the unique real solution of
(1). This situation is similar to that of the Cardan method,
though not involving the same second degree equation.
Let us continue the resolution: having found u and v,
we compute λ :
1 + λ
1

u + λv
a

= 0

⇔ (a + v)λ + (a + u)
= 0 
Equation (1) thus becomes
(

x − u
x − v 
)

^{3
}

= − λ = 
a + u
a + v

so the solution is
Note: the choice of which root of (4) is named u and the other v, is
arbitrary. In each case, the 3 third roots of λ in (6) give the 3
roots of (1) but in the opposite order.
Trigonometric version
Just like the usual Cardan solution can be expressed using the
trigonometric functions cos, ch or sh depending on cases (let us
skip this here, as it is wellknown and giving in wikipedia), the
above resolution by an homography has its own trigonometric
expression, as follows.
For the case of an equation with 3 real roots, introduce the
variable t = tan θ. By looking at the imaginary part
of (1 − it)^{3}(1 + i tan(3θ))
we get the trigonometric formula
t^{3} − 3t + (1 − 3t^{2})
tan(3θ) = 0
Equation (1) first needs to be reduced to the case c = ab
using the change of variable (5). The resulting equation
X^{3} + 3AX^{2} +
3BX + AB = 0
will be identified to the above trigonometric formula, by the system
X = α t
tan(3θ) = γ
The substitution in the above trigonometric formula multiplied by α^{3},
gives the cubic equation
X^{3} − 3α^{2}X +
γ(α^{3} − 3αX^{2}) = 0
Its identification with our equation gives
A = −αγ
B = −α^{2}
Finally
α= √−B
γ = −A/α
The condition that the equation has 3 roots, takes the form B<0.
The solution is expressed using functions tan and Arctan.
For an equation with only one simple root (B>0) we need to
use the function th or coth instead, as they satisfy the same
identity
th^{3}θ + 3thθ − (3th^{2}θ
+ 1) th(3θ) = 0
coth^{3}θ + 3cothθ − (3coth^{2}θ
+ 1) coth(3θ) = 0
with the only difference that thθ < 1 while cothθ
> 1, which gives the inequality condition that determines which
of both functions is needed.
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