Determinant fun

First we'll simplify the matrix,

$\begin{vmatrix} -bc & b^2+bc & c^2+bc \\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix} = \frac{1}{abc} \begin{vmatrix} -abc & ab^2+abc & ac^2+abc \\ a^2b+abc & -abc & bc^2+abc \\ a^2c+abc & b^2c+abc & -abc \end{vmatrix}$

$=\begin{vmatrix} -bc & ab+ac & ac+ab \\ ab+bc & -ac & bc+ab \\ ac+bc & bc+ac & -ab \end{vmatrix}$ then use $R_2 \rightarrow R_2-R_1$ first and then use $R_1 \rightarrow R_1-R_3$ and you shoud get,

$\begin{vmatrix} -ab-bc-ca & ab+bc+ca & ab+ac \\ 0 & -ab-bc-ca & ab+bc \\ ab+bc+ca & 0 & -ab \end{vmatrix} = (ab+bc+ca)^2 \begin{vmatrix} -1 & 1 & ab+ac \\ 0 & -1 & ab+bc \\ 1 & 0 & -ab \end{vmatrix}$

$=(ab+bc+ca)^3 =64 \implies ab+bc+ca=4$ (considering real solutions only!).