Prove $g(a) = b, g(b) = c, g(c) = a$ 
Let
$$f(x) = x^3 - 3x^2 + 1$$
$$g(x) = 1 - \frac{1}{x}$$
Suppose $a>b>c$ are the roots of $f(x) = 0$. Show that $g(a) = b, g(b) = c, g(c) = a$.
(Singapore-Cambridge GCSE A Level 2014, 9824/01/Q2)

I was able to prove that
$$fg(x) = -f\left(\frac{1}{x}\right)$$
after which I have completely no clue how to continue. It is possible to numerically validate the relationships, but I can't find a complete analytical solution.
 A: For all x $g^3(x)=x$ and 
$\displaystyle(x-a)(x-g(a))(x-g^2(a))=\left(x-a\right)\left(x-\frac{a-1}{a}\right)\left(x+\frac{1}{a-1}\right)=$
$=x^3-3x^2+1$.
A: Rearranging $g(x) = 1 - \frac{1}{x},$ we obtain $x = \frac{1}{1-g(x)}$. Substitute into  $f$ and we obtain $$\left(\frac{1}{1-g(x)}\right)^3 - 3\cdot\left(\frac{1}{1-g(x)}\right)^2+1.$$ The numerator of this turns out to be: $$1 - 3(1-g(x)) + (1-g(x))^3 = -(g^3(x) - 3g^2(x) +1) \equiv -f.$$ for $x\equiv g(x).$ Hence we can say that the roots of $f$ are equivalent to cyclic evaluations of $g$ for those roots, noting that any one of these roots cannot be substituted into $g$ to produce itself again since $a \neq \frac{1}{1-a} \Rightarrow a^2-a+1 \neq 0 \,  $ for real $a\, (\bigtriangleup\,=-3).$
A: 
Note: this is only a Hint

Using vieta's formulas
$$a+b+c=3$$
$$ab+bc+ca=0$$
$$g(a) = 1 - \frac{1}{a}=\frac{a-1}{a}=b\implies a-1=ab\tag{1}$$
$$g(b) = 1 - \frac{1}{b}=\frac{b-1}{b}=c\implies b-1=bc\tag{2}$$
$$g(c) = 1 - \frac{1}{c}=\frac{c-1}{c}=a\implies c-1=ca\tag{3}$$
Adding $(1),(2),(3)$
$$a+b+c-3=ab+bc+ca$$
Which is true from relation between roots given by vieta's formula
Now construct your proof
