Answer: The cubic equation is \(x^3-3x^2+4x+7=0\).

Let the required cubic have roots \(\alpha,\beta,\gamma\). Write

\(x^3+px^2+qx+r=0\).

By Vieta's formulas,

\(\alpha+\beta+\gamma=-p\),

\(\alpha\beta+\beta\gamma+\gamma\alpha=q\),

\(\alpha\beta\gamma=-r\).

We are given \(\alpha+\beta+\gamma=3\), so \(p=-3\).

Next use \(\alpha^2+\beta^2+\gamma^2=1\). Since

\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\),

we have

\(1=3^2-2q\),

so \(1=9-2q\), hence \(q=4\).

Now use \(\alpha^3+\beta^3+\gamma^3=-30\). The identity

\(\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)+3\alpha\beta\gamma\)

gives

\(-30=3^3-3\cdot 3\cdot 4+3\alpha\beta\gamma\).

So

\(-30=27-36+3\alpha\beta\gamma=-9+3\alpha\beta\gamma\),

hence \(3\alpha\beta\gamma=-21\) and \(\alpha\beta\gamma=-7\).

Therefore \(r=7\), because \(\alpha\beta\gamma=-r\).

So the cubic equation is \(x^3-3x^2+4x+7=0\).