# Why was the cubic specifically so hard to solve?

I'm a huge fan of the history of Algebra and, recently, I've noticed a bit of an oddity. Degree one equations have been known (and solved) for as long as human history. For degree two equations, we have also known the general solution since many, many years BC, since pretty much the moment people have begun to actually study and develop mathematics.

However, when it comes to the cubic, it took literal millennia for a solution to be found, only happening around 1500 AD in Italy. One would perhaps assume that the quartic would, therefore, take some considerable time to solve (maybe not millennia, since math's growth has been more or less exponential, but many years nonetheless). Not only was this not the case, the quartic was solved in the exact same timeframe as the cubic, with a couple years removed.

So my question is: why did it take so long to solve the cubic when it took so little to solve the quartic?

The solution of the cubic stuck around as an open problem for far longer than the other degree polynomial equations, and I just can't see a reason for this... Maybe the real question is "why was the quartic so easy to solve", which I am also curious about.

Does anyone have any insights as for why this is, or any sources for further reading? I've read a few Math History books, but none of them could clarify my question.

• The quartic can be always be solved by making an appropriate substitution that reduces the problem to solving only quadratics and cubics. See Ferrari’s Solution of a Quartic Equation. Nov 4, 2021 at 12:01
• I suppose it would be self-referential to claim that solving polynomials gets exponentially harder :-) Nov 4, 2021 at 12:25
• the cubic is also solved by reducing to a quadratic --- This is not the same kind of reducing that I previously mentioned regarding quartic equations. Every (numerical) solution to a quartic equation with integer coefficients can be obtained by solving a sequence of equations, each either a quadratic or a cubic, the first having integer coefficients, and each of the succeeding equations having coefficients belonging to the field generated by all the previous solutions. This is not true for cubics -- $\sqrt[3]2$ cannot be obtained by solving a finite sequence of quadratic equations this way. Nov 4, 2021 at 19:14
• Because the sort of transformations and substitutions one needs to solve the cubic were intractable without suitable symbolic notation (even if still mixed up with text) and handy rules for manipulating symbols. As a confirmation, the irreducible case of the cubic was not resolved until Viete pushed symbolism even further. However, once that was in place it didn't take much to solve the quartic beyond what was already at hand to solve the cubic. Nov 4, 2021 at 20:24
• For an example of what I mean by a sequence of quadratic equations . . . (my previous comment), see Chapter 6. Solving $x^{257} = 1$ by Quadratic Equations (pp. 31-35) in my manuscript A detailed and elementary solution to $x^{17} = 1$. In fact, in this case the coefficients of the various quadratic equations are all integer-linear combinations of solutions to previous equations. FYI, the numerical details of this has been checked several times with $1000$-significant digit computations. Nov 4, 2021 at 20:24

There is nothing special about cubic. I think the right question is: why the quadratic equations are much easier to solve than cubic and quartic? Well it's pretty obvious, is not it? Quadratic is indeed very easy to solve. There is a natural way to explain intuition behind the solution and there is nice geometrical interpretation. Cubic and quartic are just on the next level of difficulty.

If you want a more deep answer, then I can argue that one needs symbolic algebra to solve cubic and quartic, but to solve quadratic equation one doesn't actually need algebra (although today in schools we teach quadratic using algebra, Babylonians did not know algebra, but can solve problems that involves quadratic equations). It's not coincidence that symbolic algebra was developed in XVI century by the same people that contributed to solving cubic and quartic equations.

• I agree with the second paragraph of this answer but would add that there is also another reason: the general decline of science and mathematics which lasted approximately 1000 years from antiquity to renaissance. People just were not interested in this kind of activity. Nov 6, 2021 at 13:20
• Also, apart from the putative "high art" of solving polynomial equations for their own sake, indeed, quadratic equations arise much more often "in real life", and (perhaps "therefore" or "because they"?) have manifest geometric interpretations. Mar 14, 2022 at 21:56
• Late to the party, but there you go ... I have had some success in teaching the quadratic formula to students using a geometric approach: taking the $x^2 + \frac b a x$ as a rectangle chopping the rectangle in half, making a square-with-a-corner-cut-out (much easier for a contemporary pupil than that glug of letters "gnomon"). Apparently schools don't teach that any more, they just give the students the formula and tell them to learn it. Mar 18, 2022 at 8:44

The primary reason is that you have a roadblock: you absolutely need complex numbers - even for real-valued solutions - unless you go the route of using trigonometry and/or hyperbolic functions. The simplest example of the roadblock is directly connected to the ancient problem of trisecting the 60 degree angle. Write down the cubic for the cosine of 20 degrees (obtained via the triple angle formula, using the cosine of 60 degrees): $$4c^3 - 3c = \frac{1}{2}.$$ Now, use the "Cardano" formula, and you'll quickly see the problem. In fact, there's no way to express the real root to this in terms of the four arithmetic operations and square and cube roots, without using complex numbers.

The other roadblock was negative numbers! Del Ferro solved all equations of the form $$x^3 + mx = n$$ before 1500 ... but only with non-negative coefficients $$(m,n)$$. Why? Because there wasn't any notion of negative numbers around yet!

Both complex and negative numbers came onto the scene together, not separately; because once you start talking about numbers like $$-1$$, you're also going to be asking questions like what $$\sqrt{-1}$$ is.

In the early 1500's, Tartaglia said "I have a secret"; Cardano said "Cough it up" and Tartaglia bound him to a (word-of-mouth) non-disclosure, but Cardano said: "Publish it soon or I'll end-run you" (by propping up and upgrading del Ferro's solution). Cardano's publication allowed for both negative numbers and their square roots, but left the square roots of negative numbers uninterpreted.

All of that is the account given over on the Wikipedia article Cubic Equation.

A real root to $$x^3 + 3Ax + 2B = 0$$ is $$x = C - A/C$$, where $$C^3 = -B + \sqrt{A^3 + B^2}$$. So, with the cosine of 20 degrees, $$A = -1/4$$ and $$B = -1/16$$, and the cosine of 20 degrees is $$C + 1/(4C)$$, where $$C^3 = 1/16 + \sqrt{-3/256}$$ ... oops. There's your problem. What's the cube root of a complex number, expressed in the form $$x + i y$$? How do you get the $$(x,y)$$ parts for $$C$$? Well, that involves solving a cubic equation. Guess which one?

It's the Casus Irreducibilis Impasse. According to the linked Wikipedia article, the proof that it actually is an impasse was established in 1843. So, in 300+ years' retrospect: that's your roadblock. You're forced either into complex numbers (never mind negative numbers) or - in this example - into trigonometry.

I had a similar problem, when in high school. I asked my 11th grade algebra teacher what the solution to the cubic $$x^3 + x + 1 = 0$$ was, and he said "You need to learn calculus for that". So, the following week, I went back to him and said, "Ok. I bought a \\$5 calculus book from the bookstore, carefully read through all of the sections, did all the exercises, and learned the material, but I still can't solve the equation" (in closed form). Curiously, I never got any answer from him on that, since he was too busy fainting at the time, for some strange reason. About seven months later, the following year, I asked my third semester calculus assistant professor in college, and he finally coughed it up. So ... all that running around just to find out what the closed form solution to $$x^3 + x + 1 = 0$$ was. And it's not even one of those Impasse cases.

• Complex and negative numbers are needed for some quadratics too Oct 23, 2023 at 1:49
• Not irreducibly so. The Impasse is that you can't even express the solution in x + i y form at all, with "x" and "y" individually being algebraic expressions involving the four arithmetic operators and roots, never mind whether the "i" is there or not. The "i" part is necessary - even for real solutions. Such is not the case for the quadratic equation. Oct 23, 2023 at 19:55