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 6th 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