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!    
 A: 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.
A: The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other zeros are also in $\Bbb{Q}[a]$, hence they are polynomials in $a$ with rational coefficients.

We could run full Cardano on it, but I have seen this polynomial too often, so I will take a shortcut. Let's write $\zeta=e^{2\pi i/7}$ and 
$$u=\zeta+\zeta^{-1}=2\cos\frac{2\pi}7.$$
We get from binomial formula that
$$
u^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}
$$
and
$$
u^2=\zeta^2+2+\zeta^{-2}.
$$
Therefore 
$$
p(u)=u^3+u^2-2u-1=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=\zeta^{-3}\frac{\zeta^7-1}{\zeta-1}=0.
$$
So $p(x)$ is the minimal polynomial of $u=2\cos(2\pi/7)$. What about its other zeros? Galois theory tells us that the powers $\zeta^k, k=1,2,3,4,5,6$ are exactly the conjugates of $\zeta$. Therefore the conjugates of $u$ are of the form $\zeta^{k}+\zeta^{-k}=2\cos(2k\pi/7)$.
Observe that
$$
u^2-2=4\cos^2\frac{2\pi}7-2=2(2\cos^2\frac{2\pi}7-1)=2\cos\frac{4\pi}7
$$
by the formula for the cosine of a doubled angle, so $u^2-2$ is one of the other zeros of $p(x)$. I hope that it is no longer a surprise that the third root is $2\cos\dfrac{8\pi}7$. See robjohn's answer for a way of quickly writing this as a polynomial of $u$ as well. Further observe that the angle doubling trick stops here because $2\cos\dfrac{16\pi}7=2\cos\dfrac{2\pi}7$.

We do get a cyclic splitting field whenever the discriminant of an irreducible cubic in $\Bbb{Z}[x]$ is a perfect square - that part generalizes. The trickery with roots of unity and cosines is somewhat special to this polynomial. However, by the Kronecker-Weber theorem all cyclic extensions of $\Bbb{Q}$ reside inside some cyclotomic extension. In other words the roots of such cubics can be written as polynomials with rational coefficients evaluated at some root of unity.
A: Let $\beta$ be another root of the given equation, then:
$\beta^3+\beta^2-2\beta+1 = 0 = \alpha^3+\alpha^2-2\alpha+1\to (\beta^3-\alpha^3)+(\beta^2-\alpha^2)-2(\beta-\alpha)=0\to (\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2+\beta+\alpha-2)=0\to \beta^2+(\alpha+1)\beta+\alpha^2+\alpha-2=0$ since $\alpha \neq \beta$. Using quadratic formula we can find a formula for $\beta$ in term of $\alpha$. Is this fine? Note that this method assumes that all the real roots are distinct. So you should prove this first, and it is equally interesting finding for you as well. 
A: In a more general manner, if $a$ is root of $$x^3+Ax^2+Bx+C=0$$ then $$x^3+Ax^2+Bx+C=(x-a)(x^2+Px+Q)$$ grouping terms we then have $$x^2 (a+A-P)+x (a P+B-Q)+a Q+C$$ and all coefficients must be zero. So, $$P=a+A$$ $$Q=B+aP=a^2+a A+B$$ The unused equation $$aQ+C=a^3+Aa^2+Ba+C=0$$ just confirms that $a$ is a root of the initial equation.
Now, solving the quadratic $x^2+Px+Q=0$, the two other roots are then given by $$x_{1,2}=\frac{1}{2} \left(-a-A\pm\sqrt{-3 a^2-2 a A+A^2-4 B}\right)$$
