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I am trying to learn the basics of cubic equations. Will be happy for help or guidance. As suggested, I will elaborate on my attempt:

Placing $y=\frac{x+3}{3}$, gives: \begin{align} \frac{y^3}{27} - \frac{y}{2} - \frac{2}{3} = 0 \end{align} The coefficient of the linear term is negative so we can not use the trigonometric solution in the provided link.

There is another substitution $z=x-1$ which leads to \begin{align} z^3 - \frac{3z}{2} - \frac{2}{3} = 0. \end{align} Also here, the substitution $z=u+u^{-1}$ does not help.

Morad
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    This might help you https://math.stackexchange.com/questions/2371108/what-are-the-ways-to-solve-cubic-equations – A field Jul 26 '24 at 14:32
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    Look up the rational root theorem, it's a good place to start. – Amaan M Jul 26 '24 at 14:32
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    This question is similar to: Is there anything like “cubic formula”?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – The Art Of Repetition Jul 26 '24 at 14:35
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    If you want to find exact values of $x$ satisfying equation, it might be sweaty. But at least derivative tells us that it has $3$ distinct roots – Fuat Ray Jul 26 '24 at 14:35
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    This question is similar to: How to solve $8t^3-4t^2-4t+1=0$. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Dietrich Burde Jul 26 '24 at 14:43
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    The first 2 terms are six times the first two terms of $(x-1)^3$. Then fset u=x-1. This transforms the equation into a depressed cubic by getting rid of the quadratic term. Then set u=v+p/v. Should get an expression you can express as a quadratic. This is one way of deriving the cubic formula. – TurlocTheRed Jul 26 '24 at 14:50
  • I do not see how the trigonometric tricks used there can help here - Read again this post. Your cubic has three real roots. The answers in the post explains it : When a cubic equation has $3$ real roots, one uses a trigonometric substitution! – Dietrich Burde Jul 26 '24 at 15:02
  • Your attempt is false. Setting $x=3(y-1)$ cannot depress the cubic. Better follow the advice above and set $x=y+1$. – Anne Bauval Jul 26 '24 at 15:10
  • Is there a unique substitution? it seems that both work – Morad Jul 26 '24 at 15:12
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    Any rescaling doesn't bring back a quadratic term. So, you can set $x=ky+1$ for any $k\ne0$. The simplest is $k=1$. Please delete your first (wrong) attempt with $y=\frac{x+3}{3}$. – Anne Bauval Jul 26 '24 at 15:14
  • Will the further substitution $z = u+u^{-1}$ can help here? i do not see it – Morad Jul 26 '24 at 15:16
  • @AnneBauval I cannot edit it – Morad Jul 26 '24 at 15:28
  • Yes you can, you already did several times. Click on "Edit". As for $z=u+u^{-1}$, it won't help here. Apply any method of https://en.wikipedia.org/wiki/Cubic_equation Anyway, your three roots won't look nice. Did you choose that equation at random? – Anne Bauval Jul 26 '24 at 15:39
  • @AnneBauval Not random at all, it comes from a very nice problem in online optimization. What I actually need is a particular root of this eq. and not all of them – Morad Jul 27 '24 at 12:04

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