A while back I I began a discussion about deriving formulas for solving polynomial equations. We saw that linear and quadratic polynomials did not pose much of a challenge. But cubic polynomials are considerably more complex. The set-up was that we had a polynomial equation of this form:

\[

x^3+ax^2+bx+c=0

\]

We can assume that the leading coefficient is one, so long as we're OK with the idea that the other coefficients might be fractions. If we have a cubic polynomial whose leading coefficient is not one, then we can simply divide through by whatever it is.

We got as far as saying that if you carry out the change of variables

\[

x=y-\frac{a}{3}

\]

the result is an equation of the form:

\[

y^3+py=q

\]

We referred to this sort of cubic polynomial, lacking a square term, as a *reduced cubic*.

If you work through all of the algebra you will find that

\[

p=b-\frac{a^2}{3} \phantom{xx} \textrm{and} \phantom{xx}

q= -\left( \frac{2a^3}{27}-\frac{ab}{3}+c \right)

\]

The part that's important to us is the fact that *p* and *q* are polynomial functions in the coefficients of our original equation.

The question now becomes: where do we go from here? Why was it helpful to eliminate the quadratic term? The answer lies in the algebraic identity:

\[

\left( s-t \right)^3+3st(s-t)=s^3-t^3

\]

where *s* and *t* are arbitrary real numbers.

This identity is readily verified. If you remember the binomial theorem (and who doesn't!) then you can check it pretty quickly in your head. But now take a good hard look at it. It actually has the same form as our reduced cubic. Imagine, for the moment, that we could find values for *s* and *t* with the properties that

\[

s^3-t^3=q \phantom{xxx} \textrm{and} \phantom{xxx} 3st=p

\]

Then our algebraic identity becomes

\[

(s-t)^3+p(s-t)=q

\]

This implies that *(s-t)* is a solution to our reduced cubic. It follows that finding a formula for the reduced cubic is equivalent to solving our pair of equations. This is not so hard to do using standard algebraic techniques. We begin by squaring the first equation to get

\[

s^6-2s^3t^3+t^6=q^2

\]

Then we cube the second equation to get

\[

s^3t^3=\frac{p^3}{27}

\]

When this second equation is multiplied by four and added to the first equation, the result is

\[

s^6+2s^3t^3+s^6=q^2+\frac{4p^3}{27}

\]

Not impressed? Really? Well, what if I called your attention to the fact that the left-hand side is actually a perfect square? We actually have

\[

\left( s^3+t^3)^2=s^6+2s^3t^3+t^6

\]

It follows that if we take square roots on both sides of our equation then we get

\[

s^3+t^3=\sqrt{q^2+\frac{4p^3}{27}}

\]

Time to take stock. We have one equation about the sum of the cubes of *s* and *t*. We have another that addresses the differences of the cubes of *s* and *t* (scroll up a bit to refresh your memory). Adding them together gives us

\[

2s^3=q+\sqrt{q^2+\frac{4p^3}{27}}

\]

Subtracting one from the other gives us

\[

2t^3= -q + \sqrt{q^2+\frac{4p^3}{27}}

\]

Now we divide through by 2 and take cube roots to get

\[

s=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}

\]

\[

t=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}

\]

Problem solved! We have found a solution to our system of equations. And since we have that

\[

y=s-t,

\]

we have successfully expressed *x* in terms of our coefficients *p* and *q*. In other words, we have the formula we seek.

Of course, it's a pretty complicated formula. And it's even worse than it looks, since *p* and *q* were just the coefficients of the reduced cubic. They were themselves complicated functions of our original coefficients *a*, *b* and *c*. If we tried to replace *p* and *q* with *a*, *b* and *c* in our formula above, we would be left with something *very* unwieldy indeed. If you had an actual cubic polynomial whose roots you actually wanted to find, you would do far better with a computer algebra system than you would with the cubic formula.

But it's still pretty cool that the formula exists at all. And this particular derivation, first discovered in the sixteenth century by the Italian mathematician Tartaglia and first published by his friend/rival Cardano (it's a long story), is really very clever.

I'm sure that, having come this far, you have already moved on to wondering if there is a formula for solving the general fourth degree polynomial. Indeed there is, but that's a different post...

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I knew there was a reason we learned Newton's interpolation. ;-)

The formula for the fourth degree polynomials is even worse! We're definitely talking about things of purely theoretical interest here.

It looks like the a and b from the initial polynomial were not supposed to be the a and b you derived as functions of p and q. This made the post very confusing.

One Brow --

Sorry for the confusion. I changed the notation to avoid the problem you point to. At the moment, however, for some reason a few of the equations are not rendering properly in Firefox. Everything renders fine in Internet Explorer, however, so it is not a problem with my TeX coding. Very frustrating.

Huh.

Test1:

\[ s^6-2s^3t^3+t^6=q^2 \]

Test2:

\[

s^6-2s^3t^3+t^6=q^2

\]

Test3:

$$

s^6-2s^3t^3+t^6=q^2

$$

Test4:

$$ s^6-2s^3t^3+t^6=q^2 $$

Hypothesis:

Using 'dollarsigndollarsign' as a delimiter works for rendering the problem equations in FF,

Test5:

$$ (s-t)^3+p(s-t)=q $$

Test6:

$$ 2t^3= -q + \sqrt{q^2+\frac{4p^3}{27}} $$

Owlmirror --

Thanks for the information. It's weird that using backslash bracket works sometimes but not other times. And that other browsers don't seem to have that problem.

I wonder if it works inline?

â As we all know, $$ \sqrt{2} $$ is irrational.

How are you rendering the LaTeX?

According to the Error Console on Opera,

[07/19/2011 04:58:23 PM] JavaScript - http://mathcache.appspot.com/?svgcb=onSVGResponse&url=http://scienceblo…^3%3d+-q+%2b+%5Csqrt{q^2%2b%5Cfrac{4p^3}{27}}+%5C]

isn't loading.

I don't see how

$$x = s - t$$

It looks to me like it should be

$$y = s - t$$, and so

$$x = s - t - \frac{a}{3}$$

And since a cubic equation has 1, 2, or 3 roots, how does one extract the three roots from this formula?

Blaise Pascal --

Good catch. I've corrected the error. In happier news, the TeX now seems to be rendering properly in both Firefox and Safari.

It's actually a tricky question how the values we found for

sandtultimately translate into the roots for the original polynomial. Ultimately it comes down to the fact that every real number (other than 0) has three distinct cube roots, exactly two of which will be complex numbers. That means there is some ambiguity in the formulas we derived forsandt. But a full explanation would require a blog post of its own.Please sir! Please sir!

Actually - a related book recommendation: I have just finished reading "Why Beauty is Truth" by Ian Stewart, which traces (through the personalities involved - including the aforementioned Tartaglia and Cardano) the history of trying to solve polynomials using radicals (including, crucially, the quintic)... and from here moving on to the maths of symmetry, the development of group theory, and its impact on theoretical physics.

Fascinating book. Admirably light on hardcore algebra, but still bloody difficult for a layman like me to follow, due to the concepts involved. But ultimately (with perseverance) worthwhile - I'd happily recommend it.

Wow! I tried to work this out on my own during my sophomore year of my undergraduate studies. Got hopelessly entangled in variable transformations and needless examinations of slopes and such. It wasn't until an admirable professor in Calculus pointed me in the direction of Tartaglia that I was able to work out how to do it. But here now is the most succinct explaination of the cubic solution I've found since.

Thanks!

For more on the development of algebra, I recommend

Unknown quantity : a real and imaginary history of algebraby John Derbyshire. It is not easy reading, but it is worth plowing through. He covers the discovery of formulae for third- and fourth-degree equations in detail.Shouldn't there also be a minus sign in front of the square roots?