<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] 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
$$\{2, -1 + i\sqrt{2}, -1 - i\sqrt{2}\} $$
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