$x^{3}-x-2=0$ how do I find the roots? I can't seem to find a way to change this equation into its irreducible form to solve it:
$$x^{3}-x-2=0.$$ 
It should have 1 real root and 2 complex roots.
 A: Let $$f(x) = x^3 -x-2 $$
To solve: $$x^3 -x-2 =0$$
As you have mentioned in the problem, you are aware of the nature of the roots. That is: 2 complex and 1 real. This can be found by studying the behaviour of the discriminant. The demand is to unearth these roots.
First let us determine if this real root is positive of negative. Notice
$f(0) = -2$ and $f(2) = 4$
The function is continous and therefore must be cutting the x axis once. These values were chosen arbitrarily just to see if the REAL root is positive or negative. We have now determined: it is positive.
Subsititute $$x= t+\frac{a}{t}$$
As you said you will use this for the future, this is what you must substitute into the depressed cubic with 1 real and 2 complex roots.
“a” and “t” are both constants. “t” is a constant chosen by us in order to make the expression simpler. You will see why later.
$$x^3 -x-2 =0$$
$$(t+\frac{a}{t})^3 -(t+\frac{a}{t})-2=0 $$
Expand this one:
$$t^3+3ta+3\frac{a^2}{t} +\frac{a^3}{t^3}-t-\frac{a}{t}-2=0 $$
Let us collect like terms.
$$t^3+t(3a-1)+\frac{1}{t}(3a^2 - a)+\frac{a^3}{t^3}-2=0 $$
I want to chose such an “a” so that the middle two terms will cancel out in order for the expression to simplify.
$$(3a-1) = 0 \cap (3a^2 - a) = 0$$
$$a= 1/3$$
The expression now becomes:
$$t^3+\frac{a^3}{t^3}-2=0 $$
Multiply by $t^3$
and substitute $a=\frac{1}{3}$
$$t^6-2t^3+(\frac{1}{27})=0 $$
Notice the above expression can be made as a quadratic by substituting $t^3 = p$
$$p^2-2p+(\frac{1}{27})=0 $$
By the quadratic formula
$$p = \frac{9+\sqrt(78)}{9}$$
Substitute all back to x
Recall:
$$p = \frac{9+\sqrt(78)}{9}$$
$$t^3 = p$$
$$x= t+\frac{a}{t}$$
where $$a= 1/3$$
Here is a link to all three solutions on wolfram. If you want a method to obtain the other two complex roots comment here.
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=7953c4ea52a4873d32cc72052f3dcb10&title=Solve%20for%20X%20Calculator&theme=blue&i0=x%5E3%20-%20x%20-2&podSelect=&includepodid=Input&includepodid=Result&podstate=Result__Show%20steps&showAssumptions=1&showWarnings=1
A: Assuming that the equation is supposed to be $x^3+x-2=0$, most people would 
start with the rational root theorem.  The leading coefficient is $1$ and the constant term is $2$, so any rational root would have to be one of $\pm\frac{1}{1}$ or $\pm\frac{2}{1}$.  If you plug in $x=1$, you get $0$, so divide the left side by $(x-1)$ to get $x^2+x+2$.  Set this last equal to $0$ and use quadratic formula to find your two complex roots.
A: Its a cubic of type x³ - x - 2 = 0, To solve this let us assume x = y + z and substituting in the original equation we get (y+z)³ - (y+z) - 2 = 0, this can be rearranged further of the form 
y³ + z³ + (3yz - 1)(y + z) -  2 = 0
Put 3yz + 1 = 0 which implies z= 1/3y,
Putting this in above equation we get 
27y³ + 1/27y³ - 2 = 0
Solving this as a quadratic in y³ we get 
Solution for y and subsequently z and then required x. And while taking the roots take account of complex cube roots too to get non real root
A: More of a comment, but restricted by that format.
$$
f(x) = x^3-x-2
$$

$$
\begin{align}
  x_{re} &= \frac{1}{3} \left(\sqrt[3]{3 \left(\sqrt{78}+9\right)}+\sqrt[3]{27-3 \sqrt{78}}\right), \\[5pt]
%
  x_{\pm} &= \frac{\left(-1\pm i \sqrt{3}\right) \sqrt[3]{9-\sqrt{78}}+i \left(\sqrt{3}\mp i\right) \sqrt[3]{\sqrt{78}+9}}{2\ 3^{2/3}} 
%
\end{align}
$$
with numeric values of
$$
\begin{align}
%
 x_{re} &= 1.52138 \\
%
 x_{\pm} &= -0.76069 \pm 0.857874 i \\
%
\end{align}
$$
To see all roots, resolve the complex plane. The dark line is the contour of value $0$.

The $0$ contours are combined below, blue for Re, red for Im:

A: Let $x^3-x-r=0$
Solution:
$X = (81 r^2 - 12)^{1/2} - 9 r$
$A = X^2/12 - 1$
$F = X (27 r^2 - 4)/18$
$G = (A^3 - 27 F^2)^{1/3}$
$Q= \begin{cases}-(-1)^{1/3}G\\(-1)^{2/3}G\\G\end{cases}$
$x=X/6 + 3 F/(Q - A)$
