How to solve $x^3-x^2+2x-1=0$ Recently I took an exam where it appeared the following equation: 
$$x^3+x^2-1=0$$
On the exam we only needed to aprox. the root to the first decimal. However Wolframalpha says that there is a "computable" root. 
I tried to work it out by trigonometric substitutions and the following:
$$(x^3+x^2-1)(x^3-x^2+1)=x^6-x^4+2x^2-1$$
Now:
$$x^2\to x$$
We get $$x^3-x^2+2x-1=0$$
From where I get nothing.
Is there anyway to work out manually the roots? And whether a third degree polynomial equation which you know that has a "computable" root can be manually worked out?
Thanks in advance.

If we start with your first line:
$$x^3+x^2-1=0$$
and let $x=\frac23y-\frac13$ and multiply both sides by $\frac{27}2$ so that we end up with
$$4y^3-3y-\frac{25}2=0$$
Recall that $4y^3-3y=\cosh(3\operatorname{arccosh}(y))$ so that we have
$$\cosh(3\operatorname{arccosh}(y))=\frac{25}2$$
and solving for $y$,
$$y=\cosh\left(\frac13\operatorname{arccosh}\left(\frac{25}2\right)\right)$$
$$x=\frac23\cosh\left(\frac13\operatorname{arccosh}\left(\frac{25}2\right)\right)-\frac13\approx0.754877666$$
and this is the only real root.
