The "Cubic Formula"


Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna.

Our objective is to find a real root of the cubic equation


The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. The solution proceeds in two steps. First, the cubic equation is "depressed"; then one solves the depressed cubic.

Depressing the cubic equation.

This trick, which transforms the general cubic equation into a new cubic equation with missing x2-term is due to Nicolo Fontana Tartaglia (1500-1557). We apply the substitution


to the cubic equation, to obtain:


Multiplying out and simplifying, we obtain the "depressed" cubic

\begin{displaymath}ay^3+\left(c-\frac{b^2}{3a}\right) y+\left(d+\frac{2b^3}{27a^2}-\frac{b c}{3a}\right)=0.\end{displaymath}

Let's try this for the example


Our substitution will be x=y+5; expanding and simplifying, we obtain the depressed cubic equation


Solving the depressed cubic.

We are left with solving a depressed cubic equation of the form


How to do this had been discovered earlier by Scipione dal Ferro (1465-1526).

We will find s and t so that

3st = A (1)
s3-t3 = B. (2)

It turns out that y=s-t will be a solution of the depressed cubic. Let's check that: Replacing A, B and y as indicated transforms our equation into

(s-t)3+3st (s-t)=s3-t3.

This is true since we can simplify the left side by using the binomial formula to:


How can we find s and t satisfying (1) and (2)? Solving the first equation for s and substituting into (2) yields:


Simplifying, this turns into the "tri-quadratic" equation


which using the substitution u=t3 becomes the quadratic equation


From this, we can find a value for u by the quadratic formula, then obtain t, afterwards s and we're done.

Let's do the computation for our example


We need s and t to satisfy
3st = 6 (3)
s3-t3 = 20. (4)

Solving for s in (3) and substituting the result into (4) yields:


which multiplied by t3 becomes


Using the quadratic formula, we obtain that


We will discard the negative root, then take the cube root to obtain t:


By Equation (4),


Our solution y for the depressed cubic equation is the difference of s and t:


The solution to our original cubic equation


is given by


Concluding remarks.

I will not discuss a slight problem you can encounter, if you follow the route outlined. What problem am I talking about?

Shortly after the discovery of a method to solve the cubic equation, Lodovico Ferraria (1522-1565), a student of Cardano, found a similar method to solve the quartic equation.

This section is loosely based on a chapter in the book Journey Through Genius by William Dunham.

Exercise 1.

Show that y=2 is a solution of our depressed cubic


Then find the other two roots. Which of the roots equals our solution



Exercise 2.

Transform the cubic equation


into a depressed cubic.


Exercise 3.

Find a real root of the cubic equation in Exercise 2.

(This is for practice purposes only; to make the computations a little less messy, the root will turn out to be an integer, so one could use the Rational Zero test instead.)


[Back] [Next: The Geometry of the Cubic Formula]
[Algebra] [Trigonometry] [Complex Variables]
[Calculus] [Differential Equations] [Matrix Algebra]

EquMATHematics home page

A. Voznesesnky

Copyright © 2005-2007 All rights reserved.
Contact us