Answer: (i) With \(y=3x-1\), \(x=\dfrac{y+1}{3}\). Substitution gives \(y^3-2y-7=0\), so \(3\alpha-1\), \(3\beta-1\), \(3\gamma-1\) are its roots.
(ii) \(S_3=21\).
(iii) \(S_{-2}=\dfrac{4}{49}\).
Put \(y=3x-1\), so \(x=\dfrac{y+1}{3}\). Substituting into
\(9x^3-9x^2+x-2=0\)
gives
\(9\left(\dfrac{y+1}{3}\right)^3-9\left(\dfrac{y+1}{3}\right)^2+\dfrac{y+1}{3}-2=0.\)
Multiply by \(3\):
\((y+1)^3-3(y+1)^2+(y+1)-6=0.\)
Expanding,
\(y^3+3y^2+3y+1-3y^2-6y-3+y+1-6=0,\)
so
\(y^3-2y-7=0.\)
Thus \(3\alpha-1\), \(3\beta-1\), and \(3\gamma-1\) are the roots of this cubic.
Let these roots be \(r_1,r_2,r_3\). For \(y^3-2y-7=0\), Vieta's formulae give
\(r_1+r_2+r_3=0,\quad r_1r_2+r_2r_3+r_3r_1=-2,\quad r_1r_2r_3=7.\)
For \(S_3\), use
\(r_1^3+r_2^3+r_3^3=(r_1+r_2+r_3)^3-3(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)+3r_1r_2r_3.\)
Since \(r_1+r_2+r_3=0\),
\(S_3=3r_1r_2r_3=21.\)
For \(S_{-2}\), first find
\(S_{-1}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}=\dfrac{r_1r_2+r_2r_3+r_3r_1}{r_1r_2r_3}=-\dfrac27.\)
Also
\(S_{-1}^2=S_{-2}+2\left(\dfrac{1}{r_1r_2}+\dfrac{1}{r_2r_3}+\dfrac{1}{r_3r_1}\right).\)
Now
\(\dfrac{1}{r_1r_2}+\dfrac{1}{r_2r_3}+\dfrac{1}{r_3r_1}=\dfrac{r_1+r_2+r_3}{r_1r_2r_3}=0.\)
Therefore
\(S_{-2}=S_{-1}^2=\left(-\dfrac27\right)^2=\dfrac4{49}.\)
