# 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 at 12:01
• I suppose it would be self-referential to claim that solving polynomials gets exponentially harder :-) Nov 4 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 -- $\sqrt2$ cannot be obtained by solving a finite sequence of quadratic equations this way. Nov 4 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 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 at 20:24