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.

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 $a\, (\bigtriangleup\,=-3).$
