Answer: The roots are \(x=-1,\, \sqrt{14},\, -\sqrt{14}\) in any order.

Let the cubic with roots \(\alpha,\beta,\gamma\) be \(x^3-s_1x^2+s_2x-s_3=0\), where

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

From the first equation, \(s_1=-1\).

Also,

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

So

\(29=(-1)^2-2s_2=1-2s_2,\)

hence

\(2s_2=1-29=-28 \quad\Rightarrow\quad s_2=-14.\)

Using \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=-1\), we have

\(\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=-1.\)

Substituting \(s_2=-14\) gives

\(\frac{-14}{s_3}=-1 \quad\Rightarrow\quad s_3=14.\)

Therefore the cubic is

\(x^3-(-1)x^2+(-14)x-14=0,\)

so

\(x^3+x^2-14x-14=0.\)

Factorising by grouping:

\(x^2(x+1)-14(x+1)=(x+1)(x^2-14)=0.\)

Hence the roots are

\(x=-1,\quad x=\sqrt{14},\quad x=-\sqrt{14}.\)