First of all I wanted to clarify that this is my first post here.

I was trying to find a solution to the general depressed cubic polynomial and was able to get to the right formula but there are some steps I made which I cannot find a proof of.

First let $t^3+pt+q=0$ with $p,q \in \mathbb{Q}$ but with no rational solution. We do know that it has a real solution.

Then I'll look for a solution of the form $$t= \sqrt[3]{u_1} + \sqrt[3]{u_2} = \sqrt[3]{a + \sqrt b} + \sqrt[3]{a-\sqrt b} $$ with $a,b \in \mathbb{Q}$

Plug it into the cubic equation: $$(\sqrt[3]{u_1} + \sqrt[3]{u_2})^3+p(\sqrt[3]{u_1} + \sqrt[3]{u_2})+q = 0 $$ $$u_1 + u_2 +3\sqrt[3]{{u_1}{u_2}}(\sqrt[3]{u_1}+\sqrt[3]{u_2})+p(\sqrt[3]{u_1}+\sqrt[3]{u_2})+ q= 0$$ $$(2a+q) + (3\sqrt[3]{{u_1}{u_2}}+p)(\sqrt[3]{u_1}+\sqrt[3]{u_2})=0$$

Now notice that $2a+q \in \mathbb{Q}$ and by hypotesis since the polynomial has no rational roots, $\sqrt[3]{u_1}+\sqrt[3]{u_2} = t \notin \mathbb{Q} $

Now, what I do not understand is why $\sqrt[3]{{u_1}{u_2}} = \sqrt[3]{a^2-b} \in \mathbb{Q}$

Because of this, since $p \in \mathbb{Q}$, it means that the following two equations are satisfied: $$2a + q = 0$$ $$3\sqrt[3]{a^2-b} +p =0$$

From the first equation we get $a = - \frac{q}{2}$ And from the second one $b=a^2+\frac{p^3}{27} = \frac{q^2}{4}+\frac{p^3}{27}$

Which yields the equation of the depressed cubic: $$t=\sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} + \sqrt[3]{- \frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} $$

As stated above, my question is how one could prove that $\sqrt[3]{a^2-b} \in \mathbb{Q}$

Also, I would like to point out that this works for a cuadratic polynomial: First note that $u_1^3$ and $u_2^3$ are both roots of a cuadratic polynomial with ration coefficients, and the solution to a "reduced" cuadratic is: $$t^2+p=0 \to t=\sqrt{-p}$$ With $-p$ being the solution to the first degree polynomial equation $x+p=0$

So a natural question would be to ask wheather the solution to $t^4+at^2+bt+c$ is of the form $\sqrt[4]{\alpha_1}+\sqrt[4]{\alpha_2}+\sqrt[4]{\alpha_3}$ with $\alpha_1$, $\alpha_2$ and $\alpha_3$ all 3 real solutions to the same cubic polynomial. I'm aware that this does not hold for fifth degree polynomials.

I do not know any Galois theory, except from well known result such as the impossibility to solve some fifth degree or higher polynomials by radicals.

Any answer or comment is well apreciated. Thank you.


1 Answer 1


For background, see the depressed cubic section of this Wikipedia Article.

It is easier for me to explain, using my own syntax, than using your syntax.

Consider the following depressed cubic equation:

$$t^3 + pt + q = 0 ~: ~p,q \in \Bbb{R}. \tag1 $$

Regardless of whether the equation in (1) above has a rational root or not, and regardless of whether the equation has 3 real roots, or only 1 real root, the basic line of attack is that you can :

  • set $~(a + b)~$ equal to $~t.$

  • simulaneously, arbitrarily require that $~(ab) = -\dfrac{p}{3}.$

That is, there should be some Complex numbers $~a,b~$ that satisfy both of the above bullet points.

So, you are assuming that such a satisfying pair of numbers, $~a,b~$ exists.

Then, you have that

$$t^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = a^3 + b^3 - pt.$$

This means that because of the arbitrary assumption that such satisfying values $~a,b~$ exist, you have that the equation in (1) above is converted to

$$a^3 + b^3 + q = 0. \tag2 $$

The whole point of this is that since $~p,q~$ are known constants, and since $~b = \dfrac{-p}{3a},~$ the equation in (2) above can be converted into an (easier to solve) quadratic equation in $~a^3 ~$:

$$a^3 + \left( ~\frac{-p}{3a} ~\right)^3 + q = 0 \implies $$

$$\left( ~a^3 ~\right)^2 - \left( ~\frac{p}{3} ~\right)^3 + q \left( ~a^3 ~\right) = 0.$$

As far as your other question:

Why is there a general solution to the 4th degree polynomial equation, but no such general solution ot the 5th degree polynomial equation?

Personally, over the years, I have asked Mathematicians for a (very) simple intuitive explanation, and have always come up empty. I have not yet studied Galois Theory. I suspect that there is no simple intuitive explanation that does not involve Galois Theory.

This means that all that you can do is buy and attack a Galois Theory book. Alternatively, the issue in the previous paragraph is (instead) often covered in Abstract Algebra books.

  • $\begingroup$ Thanks a lot! It does clarify my doubt. $\endgroup$
    – asd
    Commented Mar 27, 2023 at 1:59
  • $\begingroup$ Among the plausible elementary arguments for quintics and beyond is that Lagrange's method which works nicely for cubics and quartics fails for higher degrees. From the wikipedia page: "this works well for every degree, but, in degrees higher than four, the resulting polynomial that has the $s_i$ as roots has a degree higher than that of the initial polynomial, and is therefore unhelpful for solving". $\endgroup$
    – dxiv
    Commented Mar 27, 2023 at 2:16

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