Answer: (i) \(\alpha^{2}+\beta^{2}+\gamma^{2}=-1\)

(ii) \(\alpha^{3}+\beta^{3}+\gamma^{3}=-58\)

For the cubic \(x^{3}-5x^{2}+13x-4=0\), Vieta’s formulas give

\(\alpha+\beta+\gamma=5\), \(\alpha\beta+\beta\gamma+\gamma\alpha=13\), and \(\alpha\beta\gamma=4\).

(i) Use the identity

\(\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^{2}-2(\alpha\beta+\beta\gamma+\gamma\alpha).\)

So

\(\alpha^{2}+\beta^{2}+\gamma^{2}=5^{2}-2(13)=25-26=-1.\)

(ii) Since each root satisfies the equation, for \(x=\alpha\) we have

\(\alpha^{3}-5\alpha^{2}+13\alpha-4=0\), so

\(\alpha^{3}=5\alpha^{2}-13\alpha+4.\)

Similarly,

\(\beta^{3}=5\beta^{2}-13\beta+4\) and \(\gamma^{3}=5\gamma^{2}-13\gamma+4\).

Adding these gives

\(\alpha^{3}+\beta^{3}+\gamma^{3}=5(\alpha^{2}+\beta^{2}+\gamma^{2})-13(\alpha+\beta+\gamma)+12.\)

Substitute the values found above:

\(\alpha^{3}+\beta^{3}+\gamma^{3}=5(-1)-13(5)+12=-5-65+12=-58.\)