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 $(1)$?
 A: Guess the solution $x = 2$ by using the Rational Root Theorem. By synthetic division, we have
$$x^{3} - x - 6 = (x - 2)(x^{2} + 2x + 3)$$
Using the quadratic equation on the second equation, we obtain the solution set
$$\{2, -1 + i\sqrt{2}, -1 - i\sqrt{2}\} $$
A: You can try with candidates for integer roots, that are $\pm1,\pm2,\pm3$ and $\pm 6$. If non of them works then you still have the Cardano formulas.
A: 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.
A: Other answers give ways to get the roots, but here is where you went wrong.
You assumed $x=y-2$ to get one equation for $y$, then $x=y+3$ to get another equation.  Either is correct by itself, but you can't use both simultaneously because you don't get both $x=y-2$ and $x=y+3$ for any values of $x$ and $y$.  So you cannot assume both equations are valid and thus cannot take the difference between them to get a lower degree equation for $y$.
As others explain, $x=2$ is the only real root.
