$f(a)=b,f(b)=c,f(c)=a$, find $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ 

Let $f(x)=x^2-x-2$. It is given that $a, b,c \in \mathbb{R}$ such that
$$f(a)=b,f(b)=c,f(c)=a$$ Then find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.


My attempt:
We have:
$$\begin{array}{l}
a^{2}-a-2=b \\
b^{2}-b-2=c \\
c^{2}-c-2=a
\end{array}$$
Subtracting pair wise we get:
$$(a-b)(a+b)=a-c$$
$$(b-c)(b+c)=b-a$$
$$(c-a)(c+a)=c-b$$
Trivially $a=b=c$ satisfies. So we get $a=b=c=\sqrt{3}+1, 1-\sqrt{3}$
Thus we get
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3\sqrt{3}+3}{2}, \frac{-3-3\sqrt{3}}{2}$$
Now if $a\ne b \ne c$
Then we have:
$$(a+b)(b+c)(c+a)=-1$$
Letting $p=a+b+c$ we have
$$(p-a)(p-b)(p-c)=-1$$ $\implies$
$$p(ab+bc+ca)-abc+1=0$$
Any hint here?
 A: From the condition, we need $f\circ f\circ f(a)=a$.  Now
$$f(x) = x^2-x-2=(x-2)(x+1)$$
$$f\circ f(x) = (x^2-x-4)(x^2-x-1)=x^4-2x^3-4x^2+5x+4$$
$$\implies f\circ f\circ f(x)=(x^4-2x^3-4x^2+4x+2)(x^4-2x^3-4x^2+4x+5)$$
$$\implies f\circ f\circ f(a)-a=a^8-4a^7-4a^6+26a^5+3a^4-54a^3-3a^2+34a+10 \\=(a^2-2a-2)(a^3-3a-1)(a^3-2a^2-3a+5)$$
It is easy to note all roots are real (using Descartes and evaluation at $a=-1, 0, 1$), so we have $8$ possibilities for $a$, which fall into two cases.
Case 1: The first/quadratic factor gives $a=1\pm\sqrt3$, each of which is also a solution to $f(a)=a$, so as you have already noted, we have two values for $\frac1a+\frac1b+\frac1c=\frac3a$ from these, viz. $\frac12(\pm3\sqrt3-3)$.
Case 2:  Now we have no more roots of $f(a)=a$, hence the rest of the roots (six roots from the two cubic factors), must contribute to two possible sets of $\{a, b, c\}$.
Taking the first cubic factor,
$\alpha^3-3\alpha-1=0 \implies f(\alpha)^3-3f(\alpha)-1 = (\alpha^2-\alpha-2)^3-3(\alpha^2-\alpha-2)-1 \\=(\alpha^3-3\alpha-1)(\alpha^3-3\alpha^2+3)=0$
i.e. if $\alpha$ is a root of the first cubic, so is $f(\alpha)$.  Thus the roots of that cubic contribute one set of $\{a, b, c\}$.  Now using Vieta, sum of the reciprocal of roots is $$\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}=-3$$
Similarly the roots of the last cubic form the last set of possible $a, b, c$, hence have a sum or reciprocals $=\frac35$.
Summarising the value of $\frac1a+\frac1b+\frac1c \in \{\frac12(\pm3\sqrt3-3), -3, \frac35\}$, four possibilities.
A: The following uses the symmetry of the problem and "lucky" cancellations of terms to get the result with somewhat less brute work.
The case $a=b=c$ is obvious and will be left out.
Otherwise, with $u=a+b+c, v=ab+bc+ca, w=abc$, the OP showed that:
$$
uv - w + 1 = 0 \tag{1}
$$
A second relation can be derived by summing the three equalities $a^2-a-2-b=0$ and using the Newton identity $a^2+b^2+c^2=u^2-2v\,$:
$$
0 = u^2 - 2u - 2v - 6 \tag{2}
$$
Then, multiplying $a^2-a-2-b=0$ by $a$ and eliminating the $a^2=a+2+b$ term gives:
$$
\begin{align}
0 &= a^3 - a^2 - 2a - ab
\\ &= a^3 - (a + 2 + b) - 2 a - a b
\\ &= a^3 - 3a - b - a b - 2 \tag{3}
\end{align}
$$
Summing the three equalities and using another Newton identity $a^3+b^3+c^3=u^3 - 3uv + 3 w$:
$$
\begin{align}
0 &= (u^3-3uv+3w) - 4 u - v - 6
\\ &=  u^3 - 4 u - v - 3 uv + 3 w - 6\tag{4}
\end{align}
$$
At this point $(1),(2),(4)$ are three equations that can be resolved for $u,v,w$.
Substituting $uv = w - 1$ from $(1)$ into $(4)$:
$$
\require{cancel}
\begin{align}
0 &= u^3 - 4 u - v - 3 (\cancel{w} - 1) + \cancel{3 w} - 6
\\ &= u^3 - 4u - v - 3\tag{5}
\end{align}
$$
Eliminating $v$ between $(2)$ and $(5)$ by subtracting $2 \times (5)-(2)\,$:
$$
\begin{align}
0 &= 2 u^3 - 8 u - \bcancel{2 v} - \cancel{6} - (u^2 - 2u - \bcancel{2v} - \cancel{6})
\\ &= 2 u^3 - u^2 - 6 u
\\ &= u(2u+3)(u-2)\tag{6}
\end{align}
$$
For each $u \in \{-\frac{3}{2}, 0, 2\}$, the other variables $v,w$ follow from $(2)$ then $(4)$:
$$
(u,v,w) \in \left\{\; \left(-\frac{3}{2}, -\frac{3}{8}, \frac{25}{16}\right), \left(0, -3, 1\right), \left(2, -3, -5\right) \;\right\}
$$
The first triplet corresponds to the cubic $16 z^3 + 24 z^2 - 6 z - 25$ which has two complex conjugate non-real roots, but the other two triplets qualify, so in the end $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}= \dfrac{v}{w} \in \left\{ -3, \dfrac{3}{5}\right\}\,$ (again, not counting the two cases where $a=b=c$ are all equal to either root of the original quadratic).

[ EDIT ]   The steps above are not reversible, so the possibility exists that extraneous solutions were introduced along the way. Therefore the two solutions $(u,v,w) \in \{ \left(0, -3, 1\right), \left(2, -3, -5\right) \}$ still need to be verified to work, as was done towards the end of @Macavity's answer for the same two cubics, where both sets of roots turned out to be eligible solutions.
