**Cubic
Equations and Substitutions**

Cecil Andrew Ellard

The purpose of this note is to share some ideas about solving cubic equations. The main idea is that by looking for simple substitutions, one is led to the well-known solution of the cubic by Vieta's substitution.

Suppose we want to find the solutions to a general cubic
equation, .
We might begin with the effects of a *linear* substitution,
,
with :

, with , or

**Observation 1**: Since
,
this has degree 3.

**Observation 2**: If we can choose
such
that ,
then we can eliminate the term.

**Observation 3**: If
satisfies
,
then we can eliminate the term.

**Observation 4**: To eliminate both the
term
and the term,
and put the equation into the very solvable form
,
we would need to satisfy both and
,
which implies ,
which is not true for an arbitrary cubic equation, so we can not expect a linear
substitution to transform a general cubic into this very solvable form.

By observation 2, we can always eliminate the term, so to simplify the calculations we'll assume that the term has been eliminated, so that our cubic equation has the form .

Since our linear substitution can't reduce the degree, and
can't in general produce the form ,
as a next attempt we might try to see if a linear substitution
can
be found that would produce *another* form which could be solved by taking
cube roots: .
However, the linear substitution reduces
this to,
so the composition of these linear substitutions is a linear substitution that
reduces to
a form ,
which we saw in Observation 4 can not be done, for an arbitrary cubic.

Since linear substitutions aren't getting us very far, we
might try a *polynomial* substitution
with
,
which gives us

. If we wish to produce a polynomial of degree 3 or less , then since the degree of this polynomial is , we must have , and so , or in other words, it must be a linear substitution, which has already been discussed.

If we wish to move beyond polynomial substitutions, as a
next step we might try a *rational* substitution. Let
a
non-constant rational function, where
and
are
polynomials. Substituting this into

gives

Where we have let . So for any solution of , is a solution of our original cubic . So we are now interested in choosing and so that = 0 can be solved.

The polynomials and and their degrees determine the degree of . Our first approach might be to try to choose and so that has degree 3 or less. Since deg F 3 max {deg , deg }, We can guarantee that the degree of will be 3 or less if both and have degree at most 1.

So takes the form where and are not both 0.

If =0,
this is simply a linear substitution, which we have already ruled out using a
algebraic argument. If =0,
it is the reciprocal of a linear substitution, and if both
and
are
non-zero, it is a "linear-fractional" substitution; both of these can be ruled
out similarly by algebraic means. In fact, we can give a single general argument
(Corollary 1 below) that none of these substitutions can always result in an
which
is either quadratic or linear. To do this, we'll make use of the following "No
Reduction" lemma, which tells us when a substitution fails to reduce the number
of zeros. In fact, it follows from the lemma that when the range of the
substitution is big enough, there will be *lots* of cubic equations whose
degrees are not reduced.

**No Reduction Lemma**: Let
be
a cubic equation with three distinct solutions, and let
where
is
*any* function. If the range of
contains
all three solutions of ,
then the resulting equation also
has three distinct solutions.

**Corollary 1**: Let be
a non-constant rational function, where
and
are
polynomials of degree at most 1. Then there are infinitely many cubic equations
for
which the substitution yields ,
where the degree of is
exactly 3.

To see that corollary 1 follows from The No Reduction Lemma, notice that the Rational function in corollary 1 has a range which excludes at most one value. So we can find infinitely many cubic equations each having three distinct solutions, all in the range of . So by the No Reduction Lemma, the resulting equation has at least three distinct solutions, so deg 3. But since and are polynomials of degree at most 1, deg . So deg =3, as claimed.

**Proof of the No Reduction Lemma:** Assume the
hypotheses of the lemma, and let ,
,
and be
the three distinct solutions of .
Since the range of contains
,
,
and ,
there exist ,
,
and such
that for
.
The are
distinct andis
a function, so the , are
distinct. Also, ,
for so
the resulting equation also
has three distinct solutions, Q.E.D.

Since limiting and to being polynomials of degree at most 1 fails, the next natural step is increasing the degrees of and . As we have seen, this may increase the degree of .

The simplest substitution to try next is one with quadratic and linear. Let , and , with and . So

As above,

So so
the degree of is
6. Assuming we can't solve a general 6^{th} degree equation, we might
try to choose and
to
force to
have a general form which we *can* solve. From several options, we follow
an idea of [1] and try to force to
be a quadratic in .
(A quadratic in would
always allow us to solve for ,
then for ,
and then for ,
using .)
If is
quadratic in ,
then the coefficient of is
0, and so .
Since we have assumed ,
we get ,
so

and The coefficient of in is , which we can make 0 with the choice of . The coefficient of in is then = , while the coefficient of is , so we can make both 0 by choosing . The simplest choice for the non-zero coefficients and is , forcing . Therefore a substitution that makes is quadratic in is. Another way to write this is , so we have discovered the

"Vieta's substitution" (see [1]). We can summarize the resulting solution in the following steps:

**Step 1:** Beginning with
,
,
with ,
and with to
transform the equation into one equivalent to the form
.

**Step 2:** From
,
make the substitution ,
to transform the equation into ,
where is
a quadratic in .

**Step 3:** Solve the quadratic
for
.

**Step 4:** Use
to
solve for .

**References**

** **

[1] Weisstein, Eric W. "Cubic Formula." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CubicFormula.html

Version 1, September 4, 2006