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Given is a cubic equation \begin{align} x^3 - 3 D x -2D=0, \end{align} where $D>1$ is an integer.

I ran it in a software and it gave me that one of the roots is

$$x^*_m = \frac{D}{\sqrt[3]{\sqrt{D^2 - D^3} + D}} + \sqrt[3]{\sqrt{D^2 - D^3} + D}.$$

I have two questions here:

  1. How to prove that $x_m^*$ solves the cubic solution?
  2. Simplify $x_m^*$ since we know it is a real number.

Attempt for 2: \begin{align} x^*_m &= \frac{D}{\sqrt[3]{\sqrt{D^2 - D^3} + D}} + \sqrt[3]{\sqrt{D^2 - D^3} + D}\\ &= \frac{D^{2/3}}{\sqrt[3]{ 1 + i\sqrt{D-1} }} + \sqrt[3]{D}\sqrt[3]{1 + i\sqrt{D - 1}} \end{align}

Morad
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  • Do you have any condition on the sign of $D$? If negative integers are allowed then only one of the roots needs to be real, whereas if $D$ is positive then all the roots are real. (The "integer" aspect of this is a bit of a false lead: If $D>1$ there are three real roots; otherwise there is one real root, perhaps with multiplicity.) – Semiclassical Aug 06 '24 at 11:18
  • Indeed, $D>1$, I will add that to the question. Is there an easy way to prove that all roots are real? @Semiclassical – Morad Aug 06 '24 at 16:23
  • Note: I should have said "If $D\geq 1$" above, since in the case $D=1$ one has roots $-1,-1,2$ which is "three real roots up to multiplicity". – Semiclassical Aug 06 '24 at 16:26
  • If $D>1$, one falls into the so-called casus irreducibilis (irreducible case) of the cubic formula. One can appeal to the trigonometric formula in that case to get explicit formulas for the three real solutions. (This was noted in a previously-deleted answer.) But if all you want is to establish existence, then this is probably overkill compared to calculus-based methods. – Semiclassical Aug 06 '24 at 16:29
  • This is the well-known Cardano-formula for cubic equations. See my answer to https://math.stackexchange.com/q/2838797/349785. – Paul Frost Aug 06 '24 at 16:57

2 Answers2

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The trick is that for the expression $ x^3 - 3Dx $, the substitution $ x = t + \frac{D}{t} $ changes it to $ t^3 + \frac{D^3}{t^3} $ hence completely eliminating the 'linear term'. With that substitution, the equation becomes a quadratic $ t^6 - 2Dt^3 + D^3 = 0 $.

Solve the quadratic to get $ t^3 = D \pm \sqrt{D^2 - D^3} $.

If $ a_+ = D + \sqrt{D^2 - D^3} $ and $ a_- = D - \sqrt{D^2-D^3} $ then observe that $ a_+ a_- = D^3 $. The roots of the equations $ t^3 = a_+ $ and $ t^3 = a_- $ are of the form $ \alpha, \omega \alpha, \omega^2 \alpha $ and $ \beta, \omega \beta, \omega^2 \beta $ respectively and we have $ (\alpha \beta)^3 = a_+a_- = D^3 $ therefore we may fix $ \alpha $ and $ \beta $ so that $ \alpha \beta = D $.

Then we see that the roots of the original equation are $ \alpha + \frac{D}{\alpha}, \omega \alpha + \frac{D}{\omega \alpha} $ and $ \omega^2 \alpha + \frac{D}{\omega^2 \alpha} $. (The list $ \beta + \frac{D}{\beta}, \omega \beta + \frac{D}{\omega \beta} $ and $ \omega^2 \beta + \frac{D}{\omega^2 \beta} $ is a permutation of the previous list, where the last two roots are flipped)

One has to be careful about which cube root of $ t^3 = a_+ $ you pick for the expression of $ x_m^* $ that you write.

Cranium Clamp
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  • Wow. Does it show that the root $\alpha + \frac{D}{\alpha}$ is real? For clarity, can you please add what are omega, beta for the answer. – Morad Aug 06 '24 at 09:01
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    The quadratic equation gives you two possible values for $ t^3 $, I labelled the one with the plus sign as $ a_+ $ (and similarly $ a_- $) – Cranium Clamp Aug 06 '24 at 09:21
  • To get $ t $, we need to solve $ t^3 = a_+ $ and $ t^3 = a_- $ but the roots of the first equation are of the form $ \alpha, \omega \alpha, \omega^2 \alpha $ where $ \alpha $ is SOME root, and $ \omega $ denotes a complex cube root of $ 1 $. Similarly I used $ \beta $ to denote SOME root for the second equation and later, argued that you can choose them so that $ \alpha \beta = D $ – Cranium Clamp Aug 06 '24 at 09:24
  • All this was done to do away with $ a_- $ as that equation gives the same values of $ x $ (and your radical expression has the plus sign, so I continued with $ a_+ $) – Cranium Clamp Aug 06 '24 at 09:26
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    It does seem worth noting that there are a few different routes to the cubic formula, with the one given here being that of Vieta. (See the Wikipedia page for deriving the roots of the cubic equation for more details and some other methods.) – Semiclassical Aug 06 '24 at 11:03
  • I would also emphasize that it can be decidedly non-obvious whether the roots given by Cardano's formula are in fact real. For instance, the $D=2$ case yields the polynomial $t^6-4t^3+8=(t^3-2)^2+4=0$. This yields $t^3=2\pm 2i$ and thus complex values for $t$. Yet the roots of $x^3-6x-4=0$ are $x=-2,1+\sqrt{3},1-\sqrt{3}$. It's certainly possible to show that Cardano's formula does generate these real roots, but requires taking cube roots of complex numbers correctly. (Indeed, this is why cubics and not quadratics were the historical motivation for taking complex numbers seriously.) – Semiclassical Aug 06 '24 at 11:45
  • Thanks @Semiclassical for the remarks. I plan to add my calculations on these roots. – Morad Aug 06 '24 at 16:28
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To focus on the "existence" side of the problem, here is a proof based on the intermediate value theorem that $f(x)=x^3-3Dx-2D$ has three distinct real roots when $D>1$. (I will take for granted that polynomials are continuous functions and thus IVT applies.) Note that $\sqrt{D}$ is well-defined under this condition, so we may evaluate $f(x)$ at $x=\sqrt{D}>0$:

$$f(\sqrt{D})=-2D\sqrt{D}-2D<0$$ Less trivially, we also have $$f(-\sqrt{D})=2D \sqrt{D}-2D=2D(\sqrt{D}-1)>0$$ Hence the intermediate value theorem indicates a root in $(-\sqrt{D},\sqrt{D})$. We could now appeal to $f(x)\to \pm \infty$ as $x\to \pm \infty$ to establish other roots, but instead note that $$f(2\sqrt{D})=8D\sqrt{D}-6D\sqrt{D}-2D=2D\sqrt{D}-2D=f(-\sqrt{D})<0$$ and similarly $f(-2\sqrt{D})=f(\sqrt{D})>0$. Hence the intermediate value theorem yields real roots in the intervals $(-2\sqrt{D},-\sqrt{D})$, $(\sqrt{D},2\sqrt{D})$ as well. Hence there are at least three real roots, and since a degree-$n$ polynomial has at most $n$ roots we conclude that these are the only ones.

Semiclassical
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