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.
Best Answer
Let
$$p(x)=x^3+3x^2-6x+1=0$$
I propose two solutions:
Observe first that function $f$ defined by $f(x)=\dfrac{1}{1-x}$ generates the following 3-elements group of homographies:
$$f(x), \ \ \ \ f^{(2)}(x):=f(f(x))=\dfrac{x-1}{x}=f^{-1}(x), \ \ \ \ f^{(3)}(x):=f(f(f(x)))=x, $$
itself a subgroup of the cross-ratio group (isomorphic to the 6-element symmetric group $S_3$) described in my answer here.
The discriminant of polynomial $p$ is
$$729=27^2$$
In such a case (see this document), the Galois group associated to polynomial $p$ is the 3-elements alternate group $A_3$. Therefore, as there is a unique (cyclic) structure for a 3-elements group, these two groups must coincide.
Let
$$u:=a; \ \ \ \ v:=\dfrac{1}{1-a}; \ \ \ \ w:=\dfrac{a-1}{a} \tag{0}$$
$$u+v+w+3=\dfrac{a^3 + 3a^2 - 6a + 1}{a(a - 1)} \tag{1}$$
which is $0$ because $a$ is a root of polynomial $p$.
$$uv+uw+vw+6=\dfrac{a^3 + 3a^2 - 6a + 1}{a(a - 1)} \tag{2}$$
equal to $0$ for the same reason.
$$uvw+1=0 \tag{3}$$
(1),(2) and (3) being equivalent to Vieta's relationships
$$u+v+w=-3, \ \ uv+vw+wu=-6, \ \ uvw=-1$$
attached to polynomial $p$, and by unicity of this correspondence, we are able to conclude that $u,v,w$ given by (0) are the roots of $p$.