None of the methods I know to factor cubics are working here ....

Question

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 ?

Answer 1

This polynomial factors as $$(x - \alpha)(x - (1 - 1/\alpha))(x - (-\alpha^2 - \alpha + 2))$$ where $\alpha$ is any root of the cubic.

You can find this by performing the long division $(x^3 - 3x + 1)/(x-\alpha)$ to get $x^2 + \alpha x + (\alpha^2 - 3)$ and then factoring that quadratic.

But we know that $1 - 1/\alpha$ is also a root so we can just long divide that out to get the final factor.

Answer 2

If $\alpha$ is a root of the cubic, then so is $1-\dfrac{1}{\alpha}$.

Therefore, the third root is $1-\dfrac{1}{1-\dfrac{1}{\alpha}}=\dfrac{1}{1-\alpha}$.

Alternatively, since the product of the three roots is $-1$, third root is $\dfrac{-1}{\alpha\left(1-\dfrac{1}{\alpha}\right)}=\dfrac{1}{1-\alpha}$.

Finally, since $1 = (x^3-3x+1)+(-x^2 - x + 2)(1-x)$, we have $\dfrac{1}{1-\alpha}=-\alpha^2 - \alpha + 2$.

Bottom line, the splitting field of $f(x)=x^3-3x+1$ is $\mathbb Q(\alpha)$, where $\alpha$ is any root of $f$.