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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.

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  • $\begingroup$ Neat question. This is probably a bit high powered, but I wonder if there is a Galois theoretic approach. $f(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein at $3$ for $f(x+2)$. Further, the discriminant of $f(x)$ is $81$, a square in $\mathbb{Q}$, so the Galois group of the splitting field of $f(x)$ over $\mathbb{Q}$ is $A_{3}$, i.e. cyclic of order $3$. I hit a road block here in showing that $g(x) = 1-x^{-1}$ must permute the roots in the desired fashion - would be interested to see if anyone can finish this approach, if it even can work at all! $\endgroup$ Dec 1 '14 at 4:08
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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 $a\, (\bigtriangleup\,=-3).$

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For all x $g^3(x)=x$ and $\displaystyle(x-a)(x-g(a))(x-g^2(a))=\left(x-a\right)\left(x-\frac{a-1}{a}\right)\left(x+\frac{1}{a-1}\right)=$ $=x^3-3x^2+1$.

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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

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  • $\begingroup$ Minor error: $abc = -1$ $\endgroup$
    – Hernandez
    Dec 1 '14 at 4:06

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