Express roots in polynomials of equation $x^3+x^2-2x-1=0$ 
If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.

I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-power terms, etc).
In the comment in this link, someone suggest that If $A$ is a root of $x^6−2x^5+3x^3−2x−1=0$, then so is $−A^5+2A^4−3A$, without further explanation (or maybe it's obvious to math experts, not to me) but I'm more interested in the underlying theory, preferably elementary, and techniques to solve problems of this kind.
Thanks!    

If $a$ is one root, then we get
$$
\frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1}
$$
Using the quadratic equation to solve this yields
$$
\frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2}
$$
for the other two roots.

After the Question Change
Changing the constant term only changes $(1)$ slightly:
$$
\frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3}
$$
and we still have $(2)$.
Since the discriminant is $9-2a-3a^2$, we know that if one root is between $\frac{-1-2\sqrt{7}}3$ and $\frac{-1+2\sqrt{7}}3$, then all three roots are. If any root is outside that interval, the other two roots will not be real.

After considering a comment by KCd, I see that
$$
(x-a)(x-a^2+2)(x+a^2+a-1)\equiv x^3+x^2-2x-1\pmod{a^3+a^2-2a-1}
$$
Therefore, if $a$ is a root, then $a^2-2$ and $-a^2-a+1$ are also roots.
