Answer: (i) The equation with roots \(\alpha^3,\beta^3,\gamma^3\) is \(y^3+3y^2+2y+1=0\), so \(\alpha^3+\beta^3+\gamma^3=-3\).

(ii) \(S_{-3}=-2\).

(iii) \(S_6=5\) and \(S_9=-12\).

(i) Let \(y=x^3\). Since \(x^3-x+1=0\), we have

\(y-x+1=0\), so \(x=y+1\).

But \(x=y^{1/3}\), hence

\(y^{1/3}=y+1\).

Cubing both sides gives

\(y=(y+1)^3=y^3+3y^2+3y+1\),

so

\(y^3+3y^2+2y+1=0\).

Therefore \(\alpha^3,\beta^3,\gamma^3\) are the roots of this equation. By Vieta's formula,

\(\alpha^3+\beta^3+\gamma^3=-3\).

(ii) Put \(A=\alpha^3\), \(B=\beta^3\), \(C=\gamma^3\). For the cubic \(y^3+3y^2+2y+1=0\),

\(A+B+C=-3\), \(AB+BC+CA=2\), and \(ABC=-1\).

Hence

\(S_{-3}=\alpha^{-3}+\beta^{-3}+\gamma^{-3}=\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\dfrac{AB+BC+CA}{ABC}=\dfrac{2}{-1}=-2\).

(iii) Since \(A=\alpha^3\), \(B=\beta^3\), and \(C=\gamma^3\),

\(S_6=\alpha^6+\beta^6+\gamma^6=A^2+B^2+C^2\).

Therefore

\(S_6=(A+B+C)^2-2(AB+BC+CA)=(-3)^2-2(2)=5\),

as required.

For \(S_9\), use the equation satisfied by \(A,B,C\):

\(y^3+3y^2+2y+1=0\), so \(y^3=-3y^2-2y-1\).

Summing this over \(A,B,C\),

\(S_9=-3S_6-2S_3-3\).

Since \(S_6=5\) and \(S_3=A+B+C=-3\),

\(S_9=-3(5)-2(-3)-3=-12\).