Answer:

Let the roots of

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

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

\(\alpha+\beta+\gamma=-p,\qquad \alpha\beta+\beta\gamma+\gamma\alpha=q,\qquad \alpha\beta\gamma=-r.\)

It is given that \(\alpha+\beta+\gamma=15\), so

\(-p=15\quad\Rightarrow\quad p=-15.\)

Also

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

Substitute the given value:

\(83=15^{2}-2q,\)

so

\(2q=142,\qquad q=71.\)

Now use the extra condition \(\alpha\beta+\alpha\gamma=36\). Since \(\alpha\beta+\alpha\gamma=\alpha(\beta+\gamma)\),

\(\alpha(15-\alpha)=36.\)

Hence

\(\alpha^{2}-15\alpha+36=0,\)

so \(\alpha=3\) or \(\alpha=12\). The value \(\alpha=12\) is impossible because then \(\alpha^{2}=144\gt 83\). Therefore \(\alpha=3\).

Finally,

\(\beta\gamma=q-(\alpha\beta+\alpha\gamma)=71-36=35.\)

So

\(\alpha\beta\gamma=3\cdot 35=105,\qquad -r=105,\)

and therefore \(r=-105\).