Answer: (i) \(\frac19\)

(ii) \(\frac73\)

(iii) \(\frac29\)

A cubic equation with roots \(\frac1{\alpha\beta}\), \(\frac1{\beta\gamma}\), and \(\frac1{\gamma\alpha}\) is \(9x^3-21x^2+2x-1=0\).

For \(x^3-7x^2+2x-3=0\), Vieta's formulae give

\(\alpha+\beta+\gamma=7\), \(\alpha\beta+\beta\gamma+\gamma\alpha=2\), and \(\alpha\beta\gamma=3\).

(i) Since \((\alpha\beta)(\beta\gamma)(\gamma\alpha)=(\alpha\beta\gamma)^2\),

\(\frac1{(\alpha\beta)(\beta\gamma)(\gamma\alpha)}=\frac1{3^2}=\frac19\).

(ii)

\(\frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha}=\frac{\gamma+\alpha+\beta}{\alpha\beta\gamma}=\frac73\).

(iii)

\(\frac1{\alpha^2\beta\gamma}+\frac1{\alpha\beta^2\gamma}+\frac1{\alpha\beta\gamma^2}=\frac1{\alpha\beta\gamma}\left(\frac1\alpha+\frac1\beta+\frac1\gamma\right)\).

But \(\frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\frac23\), so the value is \(\frac13\cdot\frac23=\frac29\).

Let the required roots be \(r_1=\frac1{\alpha\beta}\), \(r_2=\frac1{\beta\gamma}\), and \(r_3=\frac1{\gamma\alpha}\). Then

\(r_1+r_2+r_3=\frac73\), \(r_1r_2+r_2r_3+r_3r_1=\frac29\), and \(r_1r_2r_3=\frac19\).

Therefore the monic equation is \(x^3-\frac73x^2+\frac29x-\frac19=0\). Multiplying by 9 gives \(9x^3-21x^2+2x-1=0\).