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I have been trying to find all the different methods for factoring cubics and so far in my search I have come across:

1)Using the sum/difference of cubes

2)The grouping method

3)Using the rational root test (and assuming you find a root) followed by synthetic division.

4)The discriminant approach ( which can be a little messy )

But I was looking over an old assignment and there was this question I got wrong at the time:

Determine the splitting field of

$f(x)=x^3-3x+1$ over $\Bbb Q$

Hint: If $\alpha$ is a root compute $f(1-\tfrac{1}{\alpha})$.

But none of the method I mentioned above give roots which are in agreement with the online calculator I'm using.

My questions are :

1) What method for factoring cubics can I use here ?

2) What are some other useful methods of factoring cubics I havent't mentioned here?( I hope to find an exhaustive list so I can always factor any cubic)

3) Is there any method which one can use on ${ANY}$ cubic, to find factors/roots ?

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  • $\begingroup$ Wikipedia gives the general formula for solving a cubic equation. But it's rather messy, I'm afraid. $\endgroup$ – TonyK Jun 14 at 11:26
  • $\begingroup$ @TonyK I tried that approach ( the discrimant) but I was getting complex roots when the onle calculator said that they were real, maybe I did it wrong though. Is the approach you linked supposed to be fullproof ? $\endgroup$ – excalibirr Jun 14 at 11:31
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    $\begingroup$ Well, yes. But not always very useful, as you found out. But you don't have to find the roots (they are all real, by the way); you just have to find the splitting field. The point of the hint is that if $\alpha$ is a root, then $f(1-\frac{1}{\alpha})=0$, so $1-\frac{1}{\alpha}$ is also a root. This, coupled with the fact that $\alpha=1-\frac{1}{\alpha}$ has no real solution, means the roots are of the form $\alpha,\beta,\gamma$, where $\alpha=1-\frac{1}{\beta},\beta=1-\frac{1}{\gamma},$ and $\gamma=1-\frac{1}{\alpha}$. That's as far as I can go with my limited knowledge of Galois theory. $\endgroup$ – TonyK Jun 14 at 11:44
  • $\begingroup$ You can find the same question in Dummit and Foote, Abstract Algebra, Third Edition, Page no. 618. They ask to find the splitting field in terms of $\alpha$ only. From Tony's work, it can be easily seen that the splitting field is $\mathbb Q(\alpha)$. $\endgroup$ – Thomas Shelby Jun 14 at 11:58
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    $\begingroup$ Here is an answer sheet for your problem. It gives one root as $e^{i2\pi/9}+e^{-i2\pi/9}$; the other two roots can be constructed from my comment. $\endgroup$ – TonyK Jun 14 at 12:02

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