Answer: (a) A cubic equation with roots \(\alpha^2,\beta^2,\gamma^2\) is \(36y^3-(p^2+36)y^2+(10p+9)y-25=0\).
(b)(i) \(p=-6\).
(b)(ii) \(\alpha^3+\beta^3+\gamma^3=5\).
For \(6x^3+px^2-3x-5=0\), with roots \(\alpha,\beta,\gamma\), the standard root sums are
\(\alpha+\beta+\gamma=-\frac{p}{6}\), \(\alpha\beta+\beta\gamma+\gamma\alpha=-\frac{1}{2}\), and \(\alpha\beta\gamma=\frac{5}{6}\).
(a) Let the new roots be \(\alpha^2,\beta^2,\gamma^2\). Their sum is
\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=\left(-\frac{p}{6}\right)^2-2\left(-\frac{1}{2}\right)=\frac{p^2+36}{36}\).
The sum of their pairwise products is
\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)\).
So
\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=\left(-\frac{1}{2}\right)^2-2\left(\frac{5}{6}\right)\left(-\frac{p}{6}\right)=\frac{1}{4}+\frac{5p}{18}=\frac{10p+9}{36}\).
Their product is
\(\alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=\left(\frac{5}{6}\right)^2=\frac{25}{36}\).
Hence a cubic with roots \(\alpha^2,\beta^2,\gamma^2\) is
\(y^3-\frac{p^2+36}{36}y^2+\frac{10p+9}{36}y-\frac{25}{36}=0\).
Multiplying by \(36\),
\(36y^3-(p^2+36)y^2+(10p+9)y-25=0\).
(b)(i) From the result above,
\(\alpha^2+\beta^2+\gamma^2=\frac{p^2+36}{36}\).
Also
\(2(\alpha+\beta+\gamma)=2\left(-\frac{p}{6}\right)=-\frac{p}{3}\).
Using \(\alpha^2+\beta^2+\gamma^2=2(\alpha+\beta+\gamma)\),
\(\frac{p^2+36}{36}=-\frac{p}{3}\).
Multiplying by \(36\),
\(p^2+36=-12p\), so \(p^2+12p+36=0\).
Therefore \((p+6)^2=0\), giving \(p=-6\).
(b)(ii) When \(p=-6\), the original equation is
\(6x^3-6x^2-3x-5=0\).
So each root \(r\) satisfies
\(6r^3=6r^2+3r+5\).
Applying this to \(\alpha,\beta,\gamma\) and adding gives
\(6(\alpha^3+\beta^3+\gamma^3)=6(\alpha^2+\beta^2+\gamma^2)+3(\alpha+\beta+\gamma)+15\).
For \(p=-6\), \(\alpha+\beta+\gamma=1\), and the given condition gives \(\alpha^2+\beta^2+\gamma^2=2\). Therefore
\(6(\alpha^3+\beta^3+\gamma^3)=6(2)+3(1)+15=30\).
Hence
\(\alpha^3+\beta^3+\gamma^3=5\).
