To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$

If $a=b=c=0$ then the equality holds. WLG we assume $a\neq 0$ then $$\begin{vmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\=\frac{1}{a}\begin{vmatrix} a(-bc)+b(a^2+ac)+c(a^2+ab) & b(ac+bc+ca) & c(ac+bc+ca)\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\(applying R_1 \to \frac1a(aR_1+bR_2+cR_3))\\ =\frac{1}{a}\begin{vmatrix} a(ab+bc+ca) & b(ac+bc+ca) & c(ac+bc+ca)\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\ =\frac{1}{a}(ab+bc+ca)\begin{vmatrix} a & b & c\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\ =\frac{1}{a}(ab+bc+ca)\begin{vmatrix} a & b & c\\ 0 & -(ab+bc+ac) & 0 \\ 0 & 0 & -(ab+bc+ca) \end{vmatrix}\\ (applying R_2 \to R_2-(a+c)R_1\, and\, R_3 \to R_3-(a+b)R_1)\\ =(ab+bc+ca)^3$$

I put my solution as an alternative (a little bit simpler). Write the original determinant as

$\begin{vmatrix} -bc & b(b+c) & c(b+c)\\ a(a+c) & -ac & c(a+c) \\ a(a+b) & b(a+b) & -ab \end{vmatrix}$

Multiply the first column with $b+c$ and the second one with c and bring $\frac{1}{c(b+c)}$ in front of the determinant:

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

Multiply the first row by $\frac{1}{c(b+c)}$ in front of the determinant:

$\begin{vmatrix} -b & b & 1\\ a(a+c)(b+c) & -ac^2 & c(a+c) \\ a(a+b)(b+c) & bc(a+b) & -ab \end{vmatrix}$

Zero the first row and third column. Add the second column to the first column, multiply the third column with $-b$ and add it to the first column. Take the factors $ab+ac+bc$ out of the determinant:

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

Now you may expand or continue the elimination in the same way.

The cases $c=0$ and $b+c=0$ must be solved separately (but these cases are much simpler).

$A:=\begin{vmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\enspace\enspace\enspace$ Shift: $\enspace C_2\to C_1$, $C_3\to C_2$, $C_1\to C_3\enspace$ => $\enspace B$

$B:=\begin{vmatrix} b^2+bc & c^2+bc & -bc\\ -ac & c^2+ac & a^2+ac \\ b^2+ab & -ab & a^2+ab \end{vmatrix}$

$C:=\begin{vmatrix} 0 & 0 & (ab+ac+bc)^2\\ (ab+ac+bc)^2 & 0 & 0 \\ 0& (ab+ac+bc)^2 & 0 \end{vmatrix}$

It's $|A|\cdot|B|=|A\cdot B|=|C|$ with $|A|=|B|$ (Sarrus rule without calculating, only a compare).

=> $|A|=\sqrt{|C|}=(ab+ac+bc)^3$

Addition:

In order to calculate the sign of the square root, it's sufficient to put in a value, e.g. $(a;b;c):=(1;1;0)$ .