2
$\begingroup$

For a depressed cubic equation $x^3 + px + q =0$ having exactly one real root, Cardano's formula gives the real root as $$\sqrt[3]{-\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$ Suppose we know that the cubic equation has exactly one real root which is an integer or rational number. For example, consider $f = x^3 +x -2$ which has exactly one real root 1. Cardano's formula gives the root as $$\sqrt[3]{1 +\sqrt{\frac{28}{27}}} + \sqrt[3]{1 -\sqrt{\frac{28}{27}}}.$$ I referred to this post and used the technique mentioned there to observe that $\sqrt[3]{1 +\sqrt{\frac{28}{27}}}$ can be denested as $\frac{1}{2}+\frac{\sqrt{21}}{6}$ and $$\sqrt[3]{1 -\sqrt{\frac{28}{27}}} = \frac{1}{2}-\frac{\sqrt{21}}{6}.$$ This recovers our integral root $1,$ as desired. I tried few more examples and I was able to denest the outer cubic radical in those cases also.

This brings us to my question. Suppose we have a cubic polynomial $f \in \mathbb{Q}[x]$ which is promised to have exactly one integral/rational root and other two to be complex conjugates. Then, can we always denest the outer cubic radical as shown in the above example? Is there any counterexample and if not, what is the proof/proof idea for denesting.

I refered to this paper in search of a solution but it is heavily loaded with Galois theory which is a bit overwhelming for me at this stage. However, I did start reading the basics of Galois theory - field extensions, Galois correspondence etc. and I hope that I would be able to understand the layman terms of any answer to this question which involves Galois theory.

$\endgroup$

1 Answer 1

4
$\begingroup$

If $n$ is your one real integer solution, then $$x^3+px+q=(x-n)(x^2+nx+b)$$ where $-nb=q, b-n^2=p$ and $n^2<4b,$ that last because there are real roots to $x^2+nx+b$ otherwise.

Substitute $p=b-n^2, q=nb$ into your formula, then see if you can eliminate the cube roots in the general case. You should be able to.

In particular:

$$\begin{align}D=\frac{q^2}{4}+\frac{p^3}{27} &=\frac{n^2b^2}{4} +\frac{(b-n^2)^3}{27}\\ &=\frac{27n^2b^2+4b^3-12n^2b^2+12n^4b-4n^6}{108}\\ &=\frac{4b^3+15n^2b^2+12n^4b-4n^6}{108}\\ &=\frac{(4b-n^2)(b+2n^2)^2}{108} \end{align}$$

So $$-\frac q2+\sqrt{D}=\frac{bn}2+\frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}$$

It turns out this has a simple cube root:

$$\begin{align}\frac{bn}2+\frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}&=\left(\frac{n}{2}+\frac 12\sqrt{\frac{4b-n^2}3}\right)^3 \end{align}$$

So you can always reduce the cube root.


In your case $n=1,b=2,$ and this can be written:

$$1+\frac{2}{3}\sqrt{\frac 7 3}=\left(\frac12+\frac12\sqrt{\frac73}\right)^3$$


Nothing in this result requires $n$ to be an integer. The equalities are all true regardless of what $n$ is. It even holds when there are other real roots, or if $n$ is a complex root.


For example if $p=-7,q=6,$ then $x^3-7x+6=0$ has roots $1,2,-3.$ Cardano requires us to s find the cube root of:

$$S:=\frac72 +\frac{10}{9}\sqrt{-3}$$

When $n=1, b=-6$ and you get:

$$S=\left(\frac12+\frac{1}{2}\sqrt{\frac{-25}{3}}\right)^3$$

When $n=2, b=-3,$ and

$$S=\left(1+\frac{1}{2}\sqrt{\frac{-16}{3}}\right)^3$$

When $n=-3, b=2$ and

$$S=\left(-\frac32+\frac{1}{2}\sqrt{\frac{-1}{3}}\right)^3$$

$\endgroup$
9
  • $\begingroup$ Very nice (+1). Your approach works always if there is only one real solution $n$, it need not be an integer. However, in general you will not be able to determine $n$ (and $b$) exactly so that the solution $\frac{n}{2}+\frac 12\sqrt{\frac{4b-n^2}3}$ has only theoretical value. I think the only case where it works explicitly is when we can determine $n$ by trying the divisors of $q$ (if $p,q \in \mathbb Z$). $\endgroup$
    – Paul Frost
    Commented May 20, 2021 at 17:40
  • $\begingroup$ @Thomas Andrews Thank you for the beautiful answer! I had been stuck on this for weeks. I was using computer algebra system but I was trying not so helpful things. In retrospect, I should have factored the numerator using software. Can one also take cube roots of complicated expressions as above using software or did you do it by hand? $\endgroup$ Commented May 20, 2021 at 17:43
  • 1
    $\begingroup$ @ThomasAndrews $S = -3 + \frac{10}{9}\sqrt{-3}$. For $n=1,b=-6$, by the formula in your answer we should get $S^{1/3} = \frac{1}{2}+\frac{1}{2}\sqrt{\frac{-25}{3}}$, as you state. However, $(\frac{1}{2}+\frac{1}{2}\sqrt{\frac{-25}{3}})^3 = -3 - \frac{10}{9}\sqrt{-3} = \overline{S}$. Though, we still get 1 as root when adding with the other radical, it still seems to be a discrepancy. For the next 2 cases, the values are correct with correct signs, only this first case is a bit deviant. $\endgroup$ Commented May 21, 2021 at 7:03
  • 1
    $\begingroup$ @PranavBisht The problem here is that we take the square root of a negative real number $r$. There are two values of $\sqrt{r}$, namely $i \sqrt{-r}$ and $-i\sqrt{-r}$, and it is not a priori clear which is the "correct" choice. It seems to be obvious that we can always decide for $i \sqrt{-r}$, but this is not true. See my answer to math.stackexchange.com/q/2838797 Also have a look at math.stackexchange.com/q/4124077 $\endgroup$
    – Paul Frost
    Commented May 21, 2021 at 11:40
  • 1
    $\begingroup$ @PranavBisht More precisely, we have $D = \frac{(4b-n^2)(b+2n^2)^2}{108}$ and therefore the square root gives $\sqrt D = \epsilon \frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}$ with $\epsilon = \pm 1$. Both values are possible as your examples show. $\endgroup$
    – Paul Frost
    Commented May 22, 2021 at 8:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .