Answer: (i) The cubic equation is \(9y^3-12y^2+4y-1=0\).

(ii) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac43\).

(iii) \(\frac{1}{\alpha^2\beta^2}+\frac{1}{\beta^2\gamma^2}+\frac{1}{\gamma^2\alpha^2}=\frac49\).

Let \(y=\frac{1}{x^2}\). Then \(x=\frac{1}{\sqrt y}\), taking the substitution in the form used to produce the required equation.

Substitute this into \(x^3+2x^2-3=0\):

\(\frac{1}{y\sqrt y}+\frac{2}{y}-3=0\).

Rearrange to isolate the square-root term:

\(\frac{1}{y\sqrt y}=3-\frac{2}{y}=\frac{3y-2}{y}\).

Multiplying by \(y\sqrt y\) gives

\(1=(3y-2)\sqrt y\).

Squaring both sides,

\(1=y(3y-2)^2\).

Hence

\(1=y(9y^2-12y+4)=9y^3-12y^2+4y\), so the cubic equation is

\(9y^3-12y^2+4y-1=0\).

The roots of this cubic are \(\frac{1}{\alpha^2}\), \(\frac{1}{\beta^2}\), and \(\frac{1}{\gamma^2}\).

Writing the equation in monic form gives

\(y^3-\frac43y^2+\frac49y-\frac19=0\).

By Vieta's formulae, the sum of the roots is \(\frac43\), so

\(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac43\).

The sum of pairwise products of the roots is \(\frac49\), so

\(\frac{1}{\alpha^2\beta^2}+\frac{1}{\beta^2\gamma^2}+\frac{1}{\gamma^2\alpha^2}=\frac49\).