Cubic Equations: Beauty of Algebra

Suhrid Saha
Cosmus Mathematicus
8 min readNov 10, 2017

(The following is an extended and unedited version of my article published in my school Math newsletter.)

Image Source: Shutter stock

Algebra is said to be the link between most of Mathematics, and dates back to the Ancient Babylonians and Egyptians. It was treated as a formal science for the first time by the Greeks, of whom the great Mathematician Diophantus (3rd century AD) is sometimes called the “Father of Algebra”. The study later flourished at the hands of the Indians, the Chinese, the medieval Arabs and later the Europeans of the Renaissance Era and Modern Era.

Most of Algebra is concerned with expressions, identities and equations, and it is the latter we are interested with at the moment. One of the special forms of equations are equations in the third degree, more commonly known as cubic equations. The early Greeks and Indians made great advances while dealing with these, but most of what we know of cubic equations is from 15th century onward, when the general solution to these equations was discussed in great detail. Our job is to do just that: attempt at solving a cubic equation of any kind.

What is a Cubic Equation?

Any equation of the form ax³+bx²+cx+d=0 where a,b,c,d are all real numbers such that a is not zero are considered as cubic equations. Thus the only difference between a cubic equation and a quadratic equation (equations in 2nd degree) is that extra term ax³ which really does change the equation entirely. You may realize this from the graphs we shall study later of cubic functions and try comparing them with any quadratic function using your graphing calculator.

What earlier mathematicians found strange was their inability to solve these equations, primarily because it is nearly impossible to solve them using simple Euclidean geometry, unlike quadratic equations, for which the Persian Mathematician Al-Khwarizmi found a general solution to by completion of squares. His method was used later by René Descartes to arrive at:

Quadratic Formula. Image Source: Wikipedia

However if one tries to find a general solution to a cubic, she may meet dead ends because it requires prerequisite knowledge of complex numbers. Amazingly, the Persian poet Omar Khayyam came up with a method to solve a cubic equation using geometrical analysis, but it was only in the 16th century that an Italian genius named Scipione del Ferro was able to solve for cubic equations of a special type known as depressed cubics, which we shall discuss shortly.

After del Ferro’s death, it was Niccolò Tartaglia who took forward the study of cubic equations by arriving at a general solution. Another Italian named Girolamo Cardano is claimed to have pressed Tartaglia into revealing the secret method (in around 1530), and promised him that he would never publish it. Alas Cardano broke the promise, and although it still remains a controversy as to whether whose method it actually was, today the method to solve a cubic equation is known as Cardano’s method.

Cardano’s Method

Consider the general cubic ax³+bx²+cx+d=0. In order to solve it easily, lets try to take the bx² term out. Transforming the left-hand side of the equation as a function on the Cartesian plane tells us that the second and third terms are responsible for horizontal transformation of the graph. If we can convert this to the form x³+3Hx+g=0, things become much easier. This form of a cubic is called a depressed cubic equation.

Perhaps one of Del-Ferro, Cardano and Tartaglia realized this and thus transformed the cubic by replacing x with the x-coordinate of the point of inflection of the cubic (shown below).

If you know some differential calculus, maybe you’d know this is the zero of the 2nd derivative of any cubic.

(The fact that the x-coordinate was the zero of the 2nd derivative wasn’t known to either of the Italians back then since Calculus was only developed by Isaac Newton and Gottfried Leibniz later in the 17th century. So if you don’t know Calculus, it’s not a problem at all!)

Point of Inflection on cubic function graph. Graphing app used: Desmos

Thus the x-coordinate was found out to be x=(-b/3a) for any cubic. Using this they replaced x: x= y- (b/3a). When this substitution was used in ax³+bx²+cx+d=0, the equation was transformed into: y³+3Hy+G=0 where H and G are: H= (c/3a)-(b²/9a²) and G=(d/a)+(2b³/27a³)-(bc/3a²).

Let y=s-t for some numbers s and t such that H=st. This implies:

G= -(s-t)³-3st(s-t) => G=t³-s³+3st(s-t)-3st(s-t) => G= t³-s³

This may take some time to get comprehended because the step of taking y as the difference between any two numbers is just arbitrary. Perhaps somebody noticed a similarity between the depressed cubic equation and identity used to find (a+b)³ and randomly thought of this. Mathematicians therefore are artists in their own right!

Now H=st => s=H/t => G= t³- (H/t)³ => t⁶-Gt³-H³=0

This is a quadratic equation in t³ and can be solved using the quadratic formula:

Equation written using MS Word

Thus from the values of t³ and s³ we can find s and t and solve for y. Therefore, we can solve for x.

Discriminant of a Cubic

In Cardano’s Method discussed above, we haven’t really gone further to find the formula to find the solutions to the general cubic since that requires an understanding of complex numbers (If you are aware of primitive roots of unity then go ahead!). However this tells us that a cubic equation must have complex roots, also because not all cubic functions plotted on the Cartesian plane intersect the x-axis.

Consider the depressed cubic x³-3Hx-G=y

Graph of x³-Hx (for some x). Graphing app used: Desmos

So solutions to the cubic can be solved for by finding the points of intersection of the curve y=x³-Hx and the line y=2Hx+G. When the former is sketched, it would look like the graph on right.

It could look like other similar curves as well, but let’s just stick to this one.

Call curve y=x³-Hx as C and line y=2Hx+G as L. C and L meet at least once, so depressed cubic must have at least one real root.

When L passes through one of C’s stationary points or vertices, that is the maximum or minimum point (look at the point where C is about to turn or change direction in the graph), the cubic has a repeated root.

So when:

Written using MS Word

(If you do not understand the notation, which uses differential calculus, you need not worry. All it means is that the slope of the line tangent to the curve at a stationary point equals to plus or minus square root of H by 3. If you still seem unconvinced or seem interested you are welcome to explore derivatives of functions.)

So if L meets C at one of the stationary points,

Written using MS Word

This implies that if depressed cubic has three real solutions, L is bound between lines passing through the two stationary points of C. Thus,

Written using MS Word

And for C to meet L at one of the stationary points, 64H³-27G²=0

Therefore, x³-3Hx-G=0 has:

1) 1 real root and 2 complex/imaginary roots if Δ<0

2) 1 real root and 2 real, equal roots if Δ=0

3) 3 distinct real roots if Δ>0

Where Δ=64H³-27G² is the discriminant. Since the nature of the roots of the general cubic also depend on the nature of the roots of the depressed cubic, we can determine nature of roots of any cubic equation using this discriminant. This method that we used falls under a Mathematical discipline called Analytic Geometry.

An Alternate Representation

Can we represent a cubic expression as the sum of two cubes? This thought may have struck your head after you noticed a similarity between a depressed cubic and the identity for (a+b)³. Let’s try and find this out.

Written using MS Word

This implies that: y = A(x- λ)³+B(x- μ)³

Thus we have represented the cubic expression as a sum of two cubes. Also,

Written using MS Word

Therefore, we also conclude that λ and μ are the roots of the equation:

Challenge Question:

Written Using MS Word

Conclusion

Mathematics is as old as human civilisation itself. Human beings have come across so many different patterns in nature and while measuring quantities in recorded history. Cubic equations are just one of the many beautiful things we have stumbled upon and luckily we know how to solve such types of equations.

It is wonderful to know that Cardano’s method is not the only method that exists which can be used to solve a cubic equation. Rafael Bombelli, François Viète, Lodovico Ferrari and so many others found out new and innovative methods to solve cubic equations. So it is always important to look at the same problem in more than one way so that Mathematics is explored with rigor. We have only discussed a few aspects of cubic equations here. There are in fact so many research papers and articles on them that the internet is filled with, but what matters here is how much one is willing to go through to enjoy the processes involved in Mathematics.

References

Please Like (or Clap), Comment and Follow this page if you love Math like me. Also if you’re interested, please do check out Vieta’s trigonometric solution to certain types of cubic equations and Euler’s solution too. (Geniuses they are, aren’t they?)

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Suhrid Saha
Cosmus Mathematicus

IB Diploma candidate at Greenwood High International School, Bangalore