Answer: (i) \(\alpha^2+\beta^2+\gamma^2=2p\).
(ii) \(\alpha^3+\beta^3+\gamma^3=3q\).
(iii) \(6(\alpha^5+\beta^5+\gamma^5)=5(\alpha^3+\beta^3+\gamma^3)(\alpha^2+\beta^2+\gamma^2)\).
For the cubic \(x^3-px-q=0\) with roots \(\alpha,\beta,\gamma\), Vieta's relations give
\(\alpha+\beta+\gamma=0,\\qquad \alpha\beta+\beta\gamma+\gamma\alpha=-p,\\qquad \alpha\beta\gamma=q.\)
(i) Use the identity
\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha).\)
Substituting the values above,
\(\alpha^2+\beta^2+\gamma^2=0-2(-p)=2p.\)
So \(\alpha^2+\beta^2+\gamma^2=2p\).
(ii) Expand the cube sum using the standard identity
\(\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)+3\alpha\beta\gamma.\)
Since \(\alpha+\beta+\gamma=0\), the first two terms vanish, leaving
\(\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma=3q.\)
So \(\alpha^3+\beta^3+\gamma^3=3q\).
(iii) Since each root satisfies \(r^3=pr+q\), multiply by \(r^2\) to get
\(r^5=pr^3+qr^2\).
Summing over \(r=\alpha,\beta,\gamma\),
\(\alpha^5+\beta^5+\gamma^5=p(\alpha^3+\beta^3+\gamma^3)+q(\alpha^2+\beta^2+\gamma^2).\)
Now substitute the results from (i) and (ii):
\(\alpha^5+\beta^5+\gamma^5=p(3q)+q(2p)=5pq.\)
Therefore
\(6(\alpha^5+\beta^5+\gamma^5)=30pq.\)
Also
\(5(\alpha^3+\beta^3+\gamma^3)(\alpha^2+\beta^2+\gamma^2)=5(3q)(2p)=30pq.\)
Hence
\(6(\alpha^5+\beta^5+\gamma^5)=5(\alpha^3+\beta^3+\gamma^3)(\alpha^2+\beta^2+\gamma^2).\)
