Answer: (b) \(\alpha\beta+\beta\gamma+\gamma\alpha=1\).

The cubic equation is \(x^{3}-4x^{2}+x+6=0\), and its roots are \(-1\), \(2\) and \(3\).

Let \(s_1=\alpha+\beta+\gamma\), \(s_2=\alpha\beta+\beta\gamma+\gamma\alpha\), and \(s_3=\alpha\beta\gamma\).

We are given \(s_1=4\) and \(\alpha^2+\beta^2+\gamma^2=14\). Using \((\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)\),

\(4^2=14+2s_2\), so \(16=14+2s_2\), hence \(s_2=1\).

So the monic cubic with roots \(\alpha,\beta,\gamma\) has the form

\(x^3-s_1x^2+s_2x-s_3=0\), that is, \(x^3-4x^2+x-s_3=0\).

To find \(s_3\), use the identity

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

Substitute the known values:

\(4^3=34+3(4)(1)-3s_3\).

So \(64=34+12-3s_3\), hence \(64=46-3s_3\), and therefore \(3s_3=-18\), giving \(s_3=-6\).

Thus the cubic equation is

\(x^3-4x^2+x+6=0\).

Now factorise:

\(x^3-4x^2+x+6=(x+1)(x-2)(x-3)\).

Therefore the roots are \(x=-1,\,2,\,3\).