Answer: The cubic equation with roots \(\beta\gamma,\gamma\alpha,\alpha\beta\) is

\(y^3-5y^2-9=0.\)

\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=25,\qquad \alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3=152.\)

Let the roots of

\(x^3+5x+3=0\)

be \(\alpha,\beta,\gamma\). By Vieta's formulae,

\(\alpha+\beta+\gamma=0,\qquad \alpha\beta+\beta\gamma+\gamma\alpha=5,\qquad \alpha\beta\gamma=-3.\)

The numbers \(\beta\gamma,\gamma\alpha,\alpha\beta\) are obtained by putting

\(x=-\frac3y,\)

because if \(y=\beta\gamma\), then \(y=\frac{\alpha\beta\gamma}{\alpha}=-\frac3\alpha\).

Substitute \(x=-\frac3y\) into the cubic:

\(\left(-\frac3y\right)^3+5\left(-\frac3y\right)+3=0.\)

Multiplying by \(y^3\),

\(-27-15y^2+3y^3=0.\)

Thus

\(y^3-5y^2-9=0.\)

Let the roots of this new cubic be \(p,q,r\). Then

\(p+q+r=5,\qquad pq+qr+rp=0,\qquad pqr=9.\)

Now

\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=p^2+q^2+r^2.\)

Using \(p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp)\),

\(p^2+q^2+r^2=5^2-2(0)=25.\)

Also, from \(p^3-5p^2-9=0\), each root satisfies \(p^3=5p^2+9\). Therefore

\(p^3+q^3+r^3=5(p^2+q^2+r^2)+9+9+9.\)

So

\(\alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3=5(25)+27=152.\)