<s>[INST] Solve $x^3-x-6=0$ 
Solve for x, $$x^3-x-6=0\tag1$$

let $x=y-2$
$$(y-2)^3-(y-2)-6=0\tag2$$
$$y^3-6y^2+11y-12=0\tag3$$
let $x=y+3$
$$(y+3)^3-(y+3)-6=0\tag4$$
$$y^3+9y^2+26y+18=0\tag5$$
$(5)-(4)$:
$$15y^2+15y+30=0\tag6$$
$$y^2+y+2=0\tag7$$
This approach, I have tried, it is not working.
How can I solve $(1)$?
 [/INST] Well, you are lucky.
$x^3 - x - 6 = 0\implies$
$x^3 - x = 6\implies$
$x(x^2 - 1) = x(x+1)(x-1)=6$ and it just happens that $6 = 1*2*3$ so if you set in order $\{(x-1),x, (x+1)\} = \{1, 2,3\}$ you get consistent results that $x -1 = 1; x=2; x+1 = 3$ give consistent results and $x = 2$.
However you need to be aware that this will almost NEVER work and you were just lucky.
In general, you can use the rational root theorem that as the leading term is $x^3$ with coefficient $1$ and the constant term is $-6$.  Then if there is a rational root it will be $\frac ab$ were $a|6$ and $b|1$.
So that will be $\pm 1, \pm 2, \pm 3; \pm 6$.
If you try them one after another you find 
$2^3 - 2 - 6 = 0$ so $x = 2$ is one solution.  So we can factor out $(x-2)$ to get 
$x^3  - x -6 = x^2(x-2) + 2x^2 - x -6=$
$x^2(x-2) + 2x(x-2) + 4x - x - 6 = x^2(x-2) + 2x(x-2) + 3x - 6=$
$x^2(x-2) + 2x(x-2) + 3(x-2) = (x-2)(x^2 + 2x + 3)$
You can try the rational root thereom on $x^2 + 2x + 3$ and get that if there is a rational root it is $\pm 1, \pm 3$.  But seeing as those weren't roots of $x^3 - x -6$ then won't be roots of $x^3 - x -6 = (x-2)(x^2 +2x + 3)$ so there aren't any other rational roots.
So we must find irrational roots with the quadratic equation or by completing the square.
To solve $x^2 + 2x + 3 = 0$ we have
$x^2 + 2x = -3$
$x^2 + 2x + 1 = -3 +1 = -2$
$(x + 1)^2 = -2$.  But there are not solutions to that as $(x+1)^2$ can't be negative.
So $x =2$ is the only solutions.
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