Answer: The cubic equation is \(t^{3}+6t^{2}-t-30=0\).

The roots are \(t=2\), \(t=-3\), and \(t=-5\).

Let the cubic with roots \(\alpha,\beta,\gamma\) be \(t^{3}-St^{2}+Pt-Q=0\), where

\(S=\alpha+\beta+\gamma\), \(P=\alpha\beta+\beta\gamma+\gamma\alpha\), and \(Q=\alpha\beta\gamma\).

We are given \(\alpha+\beta+\gamma=-6\), so \(S=-6\), and \(\alpha\beta\gamma=30\), so \(Q=30\).

To find \(P\), use

\((\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)\).

Substituting the values gives

\(36=38+2P\), hence \(P=-1\).

So the required cubic is

\(t^{3}-(-6)t^{2}+(-1)t-30=0\),

that is, \(t^{3}+6t^{2}-t-30=0\).

Now factorise. Since the roots are integers, try simple factors of 30:

\((t-2)(t+3)(t+5)=t^{3}+6t^{2}-t-30\).

Therefore the numerical values of the roots are \(2\), \(-3\), and \(-5\).