Let a,b,c be the roots of x^3-6x^2+11x-4=0. The value of the determinant beginvmatrix1+a&1&1\\1&1+b&1\\1&1&1+cendvmatrix is

Subtract the first row from the second and third rows, or expand directly: $\begin{vmatrix}1+a&1&1\\1&1+b&1\\1&1&1+c\end{vmatrix}=abc+ab+bc+ca$. (Verification: expanding gives $(1+a)(1+b)(1+c)-(1+a)-(1+b)-(1+c)+2$, which simplifies to $abc+ab+bc+ca$.) For the cubic $x^3-6x^2+11x-4$ the symmetric functions are $a+b+c=6$, $ab+bc+ca=11$ and $abc=4$. Hence the determinant equals $4+11=15$. A numerical check with the three real roots of the cubic also returns $15$. Using $a+b+c$ instead of $abc$ would give the incorrect value $17$.