Question

My last post, http://openstudy.com/study#/updates/4fa231ede4b029e9dc332a4f, showed you how to solve a quadratic equation with a substitution that is lengthy compared to the quadratic. In this post, I will show you how to solve a cubic with Cardano’s method.

Answer 1

Please get rid of that comma at the end of the hyperlink if you wish to look at the previous post.

Answer 2

We shall begin by attempting to solve the cubic by completing the cube.

Answer 3

We'll do a short review on how to solve the quadratic by completing the square. Please note, that if you don't know how to solve the quadratic, you probably will have trouble solving the cubic.
\[ax^2+bx+c=0\]
\[x^2+(b/a)x+c/a=0\]
\[x^2+2(b/(2a))+b^2/(4a^2)-b^2/(4a^2)+c/a=0\]
\[(x+b/(2a))^2=(b^2-4ac)/(4a^2)\]
\[x=(-b\pm\ \sqrt{b^2-4ac})/2a\]

Answer 4

Now, let's attempt to do the same with the cubic. For simplicity's sake, we will assume that you have already divided the 3rd degree term so its coefficient is one.
\[x^3+ax^2+bx+c=0\]
Now, the corresponding binomial expansion of the cube is
\[(x+y)^3=x^3+3x^2y+3xy^2+y^3\]
We shall attempt to write the first equation in the form of the second.

Answer 5

There seems, however, to be a very large problem. ax^2+bx only match up with 3x^3 y+3xy^2 if a=b/ There are many cases in which a=/=b, and so it appears that we are stuck in this aspect.

Now, as quoted from one of the sources I had skimmed on learning how to solve the cubic "Completing the cube is impossible for simple and compelling reasons". The reason I chose to travel a dead end was because this illustrates the difference between the cubic and the quadratic.

We shall now do a reverse analysis of the roots of a cubic equation, and come to our solution. Note that this is Gerolamo Cardano's solution, and it has a very interesting history behind it. I shall not discuss this history, because it is much too long, although one interested could do research on their own.

Answer 6

Let a+b be the root of a cubic equation. Then, the cube of this polynomial root will be \(a^3+a^2b+ab^2+b^3\). Now, let us suppose x=a+b, as well as cube and then rewrite this equation as \(x^3=a^3+b^3+ab(a+b)\).

But, by our definition, \(x=a+b\), so we may resubstitute (a+b) for x, to get
\[x^3=a^3+b^3+ab(x)\]
Now, let's apply a second substitution; \(a^3=p\) and \(b^3=q\). Then,
\[ab=\sqrt[3]{pq}\]
So, whenever we have an equation of the form \(x^3=p+q+3\sqrt[3]{pq}(x)\), we know that one of the roots is \(\sqrt[3]p+\sqrt[3]q\).

Answer 7

Now, we will start analyzing the cubic from the other direction. Our current cubic appears to be similar to the form \(x^3+fx+g=0\), but it is a little bit different. We want to solve for p and q in terms of f and g. We set up the resulting two equations:
\(p+q=g\) and \( 3\sqrt[3]{pq}=f\)
We can rewrite the second equation into a simpler form, or \(pq=(f^3)/27\)

Answer 8

Now, with some clever algebraic manipulation, we shall see how p and q were solved for.
We know that \(p+q=g\), and that \(pq=(f^3)/27\).
Square both sides of the former equation, to get \(p^2+2pq+q^2=g\).
Subtract 4pq, and then take the square root.
\(\sqrt{p^2-2pq+q^2}=\sqrt{g-(4f^3)/27}\)
In other words, \(p-q=\sqrt{g-(4f^3)/27}\).

Answer 9

Now, we can finally isolate p and q. Add the p+q equation to the p-q equation, to get
\[2p+q-q=g+\sqrt{g-(4f^3)/27}\]
or, \(p=(g+\sqrt{g-(4f^3)/27})/2\)
Subtract the second equation from the first equation to get
\((p+q)-(p-q)=g-\sqrt{g-(4f^3)/27}\)
or, \(q=(g-\sqrt{g-(4f^3)/27})/2\)

Answer 10

So, we now have the values of p and q, and given an equation in the form
\(x^3+fx+g=0\), we always know that one solution to it is a+b (which equals \(\sqrt[3]p+\sqrt[3]q\), or
\(x=\sqrt[3]{g+\sqrt{(g-(4f^3)/27)/2}}+\sqrt[3]{g-\sqrt{(g-(4f^3)/27)/2}}\)

Answer 11

Now, you probably realize at this point that we have only solved the equation when the term of second degree is eliminated; x3+ax2+bx+c=0 does not appear to be solvable. However, this is where my first link comes into play. We apply the Tschirnhaus transformation (found at http://openstudy.com/study#/updates/4fa231ede4b029e9dc332a4f ) to all cubic equations with a term of the second degree.

Answer 12

And, you might realize that we have only found one root. That is quite true. However, we can always factor the resulting quadratic (even if it is not factorisable with rational numbers); if you wish to learn how you can check [ http://assets.openstudy.com/updates/attachments/4ebeb6bfe4b0caac6ff0de73-myininaya-1321123000453-myproofnoid.pdf ]. We can finally solve the cubic equation for all three of its roots!

Answer 13

Yet, we are still not done. Most people like to have one of those big, nasty, formulas. Unfortunately, it is exceedingly time consuming to do the algebraic computation associated with finding this "large equation"; more fortunately is that people have these formulas put around on the internet. I shall now attach Wikipedia's formulas.

Answer 14

I do understand that because the background is black, it is very hard to read the text. My only suggestion would be to save the image (right-click) on the desktop, and keep it there for future (as well as offline) reference.

Answer 15

There is one final thing about the cubic. Sometimes Cardano found that he came up with numerous square roots of negative numbers in the expressions, even though the cubic must have at least one real solution - in fact, what he found surprising was that in some cases where the cubic had *three* real solutions (confirmed by factoring), it still had the square root of negative numbers. This is known as the famous "casus irreduciblis"; the irreducible case, where "irreducible" complex numbers appear in a expression that can be shown to simplify to real.

Answer 16

I must go off on a somewhat wild tangent and say that the acceptence of the existence of imaginary numbers required the Cardano's solution to the cubic. Although most people are taught that when solving the quadratic and the discriminant D is less than 0 the solution has two complex solutions (that are conjugates), mathematicians back then had no meaning for the square root of negative numbers and simply saw it as a quadratic with no solution. However, the presence of "casus irreduciblis" forced mathematicians to look into the matter, which was eventually examined in-depth by Bombelli.

Answer 17

In fact, it was not known at the time, but there are ways to transform the irreducible case into the real number it is the equivalent to. One is solving the cubic with trigonometry, which I will show with my next post. Another is using hypergeometric series, an advanced form of math that is part of calculus.

Answer 18

I have used quite a few places for my guide, but I have mainly used a translation of Euler's "Elements of Algebra". The link to the Google book is at [ http://books.google.com.tw/books?id=X8yv0sj4_1YC&dq=euler+elements+of+algebra&printsec=frontcover&source=bn&hl=en&ei=KiwwS8bdEcutlAe2waCoBw&sa=X&oi=book_result&ct=result&redir_esc=y#v=onepage&q&f=false ], which may both be found at the Wikipedia page for "Element of Algebra", as well may be downloaded as a .pdf for offline reading.

Answer 19

\[\color{blue}{\mathbb{This\ marks\ the\ end\ of\ my\ tutorial\ on\ how\ to\ solve\ the\ cubic.}}\]
\[\color{blue}{\mathbb{Thank\ you\ for\ reading,\ and\ I\ hope\ you\ enjoyed\ it!}}\]

Answer 20

If you have any questions about my guide, please feel free to ask, and I (or others) will try my best to help!

Answer 21

A final extra addition: You probably have observed that I take very little care when taking the square root. This is probably something I shouldn't have done. However, since I basically ripped many parts of this from a textbook, its accuracy should not have problems in particular.

Answer 22

@KingGeorge
@dpaInc
@nbouscal
@zzr0ck3r

Answer 23

Good work. Really. :D

Answer 24

yeah... good one inky... sounds interesting and I'm gonna try it out as soon as my knee allows me to...

Answer 25

loool