How to Simplify Roots Given by Cardano’s Formula for the Cubic Equation $x^3+x^2+x+1=0$

Question

Been trying to figure out how to simplify the first root given by this version of Cardano’s Formula: $$x_1=\sqrt[3]{\dfrac{-\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right) +\sqrt{\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right)^{2} +\dfrac{4}{27}\left(\dfrac{3ac-b^{2}}{3a^{2}}\right)^{3}}}{2}} +\sqrt[3]{\dfrac{-\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right) -\sqrt{\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right)^{2} +\dfrac{4}{27}\left(\dfrac{3ac-b^{2}}{3a^{2}}\right)^{3}}}{2}} -\dfrac{b}{3a}$$ for the cubic equation $x^3+x^2+x+1$ which has that $a=b=c=d=1$ thus yielding $$x_1=\frac{\sqrt[3]{-10+6\sqrt{3}}}{3} -\frac{\sqrt[3]{10+6\sqrt{3}}}{3} -\frac{1}{3}.$$ However, I haven’t the slightest clue how to reduce the expression down to its supposed value of $x_1=-1$ if anyone has any tips of tricks I would greatly appreciate the guidance the only thing I’ve come across is attempting to find some value(s) $w,m$ such that $(w+m\sqrt3)^3=-10+6\sqrt3$. However, this just leads to another cubic making it quite counter intuitive not to mention the fact that perhaps a perfect cube isn’t guaranteed to exists for this portion of Cardano’s formula for any and all other roots to various different cubic equations.

If anyone is also able to show how to apply AND simplify the roots $x_2$ and $x_3$ using the roots of unity, I would appreciate that. So far all I know is that $$ x_2=\omega \sqrt[3]{\dfrac{-\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right) +\sqrt{\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right)^{2} +\frac{4}{27}\left(\dfrac{3ac-b^{2}}{3a^{2}}\right)^{3}}}{2}} +\omega ^2\sqrt[3]{\dfrac{-\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right) -\sqrt{\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right)^{2} +\dfrac{4}{27}\left(\dfrac{3ac-b^{2}}{3a^{2}}\right)^{3}}}{2}} -\dfrac{b}{3a}$$ and that $$ x_3=\omega^2 \sqrt[3]{\dfrac{-\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right) +\sqrt{\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right)^{2} +\dfrac{4}{27}\left(\dfrac{3ac-b^{2}}{3a^{2}}\right)^{3}}}{2}} +\omega \sqrt[3]{\dfrac{-\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right) -\sqrt{\left(\dfrac{2b^{3} -9abc+27a^{2} d}{27a^{3}}\right)^{2} +\dfrac{4}{27}\left(\dfrac{3ac-b^{2}}{3a^{2}}\right)^{3}}}{2}} -\dfrac{b}{3a}$$ where $\omega=e^{\frac{2\pi i}{3}}=\frac{-1+i\sqrt{3}}{2}$ and $\omega^2=e^{\frac{4\pi i}{3}}=\frac{-1-i\sqrt{3}}{2}$ are the aforementioned roots of unity

I understand it’s a long, hefty problem, but it’s been driving me insane, so I need help from the community. Thanks in advance.

Answer 1

As you know, the Cardano formula is a general formula. Therefore, we can not expect simple $x=-1$ from the Cardano formula directly . Because, the simplification works if and only if, when the cubic has a rational root. If not, the Cardano's formula can not be simplified in general.

I will provide for readers $2$ powerful theorems, without proof. Please, follow the theorems.

$\color{#c00}{\textbf{Theorem 1.1 }}$

Let $(a,b)\in\Bbb Q^2$ and $\sqrt a\not\in \Bbb Q$. Then $\sqrt[3]{\sqrt a +b}=\sqrt m+n$ holds for some $(m,n)\in\Bbb Q^2$, if and only if , $P(k)$ has a rational root $k\in\Bbb Q$ and $\small{\sqrt [3]{\dfrac {b}{3k+1}}}\in\Bbb Q$.

Here $$P(k)=k^3+k^2\left(\frac {6b^2-9a}{b^2}\right)+k\left(\frac {9b^2-6a}{b^2}\right)-\frac {a}{b^2}$$

and

$$m=k\sqrt[3]{\frac{b^2}{(3k+1)^2}}$$

and

$$n=\sqrt[3]{\frac{b}{3k+1}}$$

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$\color{#c00}{\textbf{Theorem - 1.2}}$

Let $(a,b)\in\Bbb Q^2$ and $\sqrt a\not\in \Bbb Q$. Define $2n=\sqrt[3]{\sqrt a+b}-\sqrt[3]{\sqrt a-b}$. Then $n\in\Bbb Q$, if and only if , $P(k)$ has a rational root $k\in\Bbb Q$ and $\small{\sqrt [3]{\dfrac {b}{3k+1}}}\in\Bbb Q$.

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Concretely, we have :

$$\begin{align}\sqrt[3]{\sqrt a+b}-\sqrt[3]{\sqrt a-b}=2\sqrt[3]{\frac{b}{3k+1}}\end{align}$$

In your case $b=-10,\, k=3$. So,

$$\sqrt[3]{-10+6\sqrt{3}} -\sqrt[3]{10+6\sqrt{3}}=-2\,.$$

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Final words :

The above formula explicitly shows that, there is no way to avoid the cubic polynomial to calculate $k$ .

Answer 2

Let set $\begin{cases}s=\sqrt[3]{a+b}\\ t=\sqrt[3]{a-b}\end{cases}\quad$ and we are interested in $X=s+t$

Notice that:

• $s^3+t^3=2a$

• $(st)^3=(a+b)(a-b)=a^2-b^2$

Since we have the identity :

$(s+t)^3=(s^3+3s^2t+3st^2+t^3)=(s^3+t^3)+3st(s+t)$

Then we get $X^3-3\sqrt[3]{a^2-b^2}X-2a=0$

Applying for values $a=10,\ b=6\sqrt{3}$ then $$X^3+6X-20=0$$

You could say that we are not much more advanced, since we got back to a cubic equation anyway. Well if there was a rational root to begin with it will factorise.

So you calculate an approximated value of $X=1.999...$ with a calculator and see if the equation actually factors by $(x-2)$ which is indeed the case confirming the rational root.

$$(x-2)(x^2+2x+10)=0$$

The other two values being complex then $X=2$.

Answer 3

The Cardano formula provides a general method to solve cubic equations, but the general solution always contains two expressions of the form $\sqrt[3]{r \pm \sqrt s}$.

In your example you get the unpleasant solution $$x_1=-\frac{\sqrt[3]{10+\sqrt{108}}}{3} -\frac{\sqrt[3]{10-6\sqrt{108}}}{3} -\frac{1}{3}. \tag{S}$$ On the other hand, it is fairly obvious that $x_1 = -1$ is a solution of the cubic equation, and this gives $x_2 = i$ and $x_3= -i$ as the two other solutions.

So it should be possible to explicitly compute the expressions $\sqrt[3]{10\pm\sqrt{108}}$. But how can that be done?

Let us more generally consider $r,s \in \mathbb Q$ and ask whether $\sqrt[3]{r \pm\sqrt{s}}$ can be written in the form $$u + \epsilon \sqrt v$$ with $u, v \in \mathbb Q$ and $\epsilon = \pm 1$.

In When does $r + \sqrt s$ with $r,s \in \mathbb Q$ have a cubic root of the form $u \pm \sqrt v$ with $u,v \in \mathbb Q$? you can see when this can be done.

1. Check the necessary condition that $\alpha = \sqrt[3]{s - r^2} \in \mathbb Q$.

In your example we get $\alpha = 2$.

2. Check whether $x^3 + 3\alpha x - 2r = 0$ has a rational root $n$. This can easily be done by the rational root theorem.

In your example we have to consider $x^3 + 6x -20 = 0$. It is easy to see that $n = 2$ is a rational root.

3. Next take $\epsilon = \operatorname{sign}( \alpha + n^2)$.

In your example $\epsilon = +1$.

4. Then we get $\sqrt[3]{r \pm\sqrt{s}} = \frac n 2 \pm \epsilon\sqrt{\alpha + \frac{n^2}{4}}$ (see equation $(8)$ in the linked question and answer).

In your example $$\sqrt[3]{10 \pm\sqrt{108}} = \frac 2 2 \pm \sqrt{2 + 1} = 1 \pm \sqrt 3$$

Inserting into $(S)$ gives $x_1 = -1$, as expected.

Remark.

The "Cardano solution" of a cubic equation always involves two expressions of the form $\sqrt[3]{r \pm \sqrt s}$. If you have already found a rational solution via the rational root theorem, then you get an unexpected equation $$\sqrt[3]{r + \sqrt s} + \sqrt[3]{r -\sqrt s} = \rho \tag{*}$$ with some rational number $\rho$. It seems that you want to find a nice trick to simplify the expressions $\sqrt[3]{r \pm \sqrt s}$ in order to understand (*).

Unfortunately there is no way to simplify them without using the rational root theorem for a certain "derived" cubic equation. But if the rational root theorem applies to the derived cubic, then it also applies to the original cubic and you may argue that the Cardano solution is not useful in that case because it gives a complicated formula for a rational number.

However, if the original cubic does not have a rational root, then the Cardano solution is definitely useful because it gives an explicit formula (you can of course always solve cubic equations by numerical methods, but this does not give a formula in terms of coefficients).

In my answer I focused on the explicit computation of the expressions $\sqrt[3]{r \pm \sqrt s}$. We have $$\sqrt[3]{10 \pm\sqrt{108}} = 1 \pm \sqrt 3 \tag{R}$$

which was obtained by a sort of algorithm (Steps 1, 2, 3). Step 2 involves finitely many trials. If you are only interested in the solution of the original cubic, it is indeed pointless to make efforts to denest $\sqrt[3]{10 \pm\sqrt{108}}$. You are better advised to apply the rational root theorem to original cubic.

Anyway, I think it is nice to verify that $$\sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 -\sqrt{108}} = 2 $$ by explicitly denesting both terms on the LHS.

Answer 4

This is equivalent to showing that ${\sqrt[3]{-10+6\sqrt{3}}} - {\sqrt[3]{10+6\sqrt{3}}}=-2$.

Simply substitute cubes for both terms under the radicals so that $a^3=-10+6\sqrt3$ and $b^3 = 10+6\sqrt3$.

Then $a = -1+ \sqrt3$ and $b = 1+\sqrt3$ and

${\sqrt[3]{-10+6\sqrt{3}}} - {\sqrt[3]{10+6\sqrt{3}}} = {\sqrt[3]{a^3}} - {\sqrt[3]{b^3}}=a-b=(-1+ \sqrt3) - (1+\sqrt3)=-2.$