The "Cubic Formula"

Introduction.

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

ax³+bx²+cx+d=0.

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

$\begin{displaymath}x=y-\frac{b}{3a}\end{displaymath}$

to the cubic equation, to obtain:

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

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

2x³-30x²+162x-350=0.

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

y³+6y-20=0.

Solving the depressed cubic.

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

y³+Ay=B.

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:

(s³-3s²t+3st²-t³)+(3s²t-3st²)=s³-t³.

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

$\begin{displaymath}\left(\frac{A}{3t}\right)^3-t^3=B.\end{displaymath}$

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

$\begin{displaymath}t^6+Bt^3-\frac{A^3}{27}=0,\end{displaymath}$

which using the substitution u=t³ becomes the quadratic equation

$\begin{displaymath}u^2+Bu-\frac{A^3}{27}=0.\end{displaymath}$

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

y³+6y=20.

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:

$\begin{displaymath}\frac{8}{t^3}-t^3=20,\end{displaymath}$

which multiplied by t³ becomes

t⁶+20t³-8=0.

Using the quadratic formula, we obtain that

$\begin{displaymath}t^3=-10\pm\sqrt{108}.\end{displaymath}$

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

$\begin{displaymath}t=\sqrt[3]{-10+\sqrt{108}}.\end{displaymath}$

By Equation (4),

$\begin{displaymath}s^3=20+t^3=20+\left(-10+\sqrt{108}\right)=10+\sqrt{108}.\end{displaymath}$

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

$\begin{displaymath}y=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}.\end{displaymath}$

The solution to our original cubic equation

2x³-30x²+162x-350=0

is given by

$\begin{displaymath}x=y+5=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}+5.\end{displaymath}$

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

y³+6y-20=0.

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

$\begin{displaymath}y=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}?\end{displaymath}$

Answer.

Exercise 2.

Transform the cubic equation

x³-6x²+14x-15=0

into a depressed cubic.

A. Voznesesnky
98-05-20