Answer: \(\alpha^{3}+\beta^{3}+\gamma^{3}=19\).

\(\alpha^{4}+\beta^{4}+\gamma^{4}=27\).

The required cubic equation is \(5x^{3}-16x^{2}+18x-6=0\).

Let \(S_n=\alpha^n+\beta^n+\gamma^n\).

For the cubic \(z^3-z^2-z-5=0\), Vieta's relations give

\(\alpha+\beta+\gamma=1\), \(\alpha\beta+\beta\gamma+\gamma\alpha=-1\), and \(\alpha\beta\gamma=5\).

Now

\(S_2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=1^2-2(-1)=3\).

Also, using \((\alpha+\beta+\gamma)^3=\alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma\), we get

\(S_3=1^3-3(1)(-1)+3(5)=1+3+15=19\).

So \(\alpha^3+\beta^3+\gamma^3=19\).

To find \(S_4\), use the recurrence coming from \(z^3=z^2+z+5\). Multiplying by \(z^{n-3}\) shows that each root satisfies \(z^n=z^{n-1}+z^{n-2}+5z^{n-3}\) for \(n\ge 3\). Summing over \(\alpha,\beta,\gamma\) gives

\(S_n=S_{n-1}+S_{n-2}+5S_{n-3}\).

Hence

\(S_4=S_3+S_2+5S_1=19+3+5(1)=27\).

Therefore \(\alpha^4+\beta^4+\gamma^4=27\).

For the transformed roots, set

\(x=\frac{z-1}{z}=1-\frac{1}{z}\).

Then

\(\frac{1}{z}=1-x\), so \(z=\frac{1}{1-x}\).

Substitute this into \(z^3-z^2-z-5=0\):

\(\frac{1}{(1-x)^3}-\frac{1}{(1-x)^2}-\frac{1}{1-x}-5=0\).

Multiplying by \((1-x)^3\) gives

\(1-(1-x)-(1-x)^2-5(1-x)^3=0\).

Expanding and simplifying:

\(1-(1-x)-(1-2x+x^2)-5(1-3x+3x^2-x^3)=0\)

\(\Rightarrow 5x^3-16x^2+18x-6=0\).