<s>[INST] To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to directly expand the determinant before greatly simplifying it).
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\ 
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}= (ab+bc+ca)^3$$
I tried everything: 
$R_1\to R_1+R_2+R_3$ and similar transformations to extract that $ab+bc+ca$ term, but to no avail. $C_2\to C_1+C_2$ and $C_3\to C_3+C_1$ seemed to be a good lead, but I couldn't follow up.
How can I solve this question?
 [/INST] If $a=b=c=0$ then the equality holds.
WLG we assume $a\neq 0$ then
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\ 
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\=\frac{1}{a}\begin{vmatrix}
a(-bc)+b(a^2+ac)+c(a^2+ab) & b(ac+bc+ca) & c(ac+bc+ca)\\ 
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\(applying R_1 \to \frac1a(aR_1+bR_2+cR_3))\\
=\frac{1}{a}\begin{vmatrix}
a(ab+bc+ca) & b(ac+bc+ca) & c(ac+bc+ca)\\ 
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\
=\frac{1}{a}(ab+bc+ca)\begin{vmatrix}
a & b & c\\ 
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\
=\frac{1}{a}(ab+bc+ca)\begin{vmatrix}
a & b & c\\ 
0 & -(ab+bc+ac) & 0 \\
0 & 0 & -(ab+bc+ca)
\end{vmatrix}\\
(applying R_2 \to R_2-(a+c)R_1\, and\, R_3 \to R_3-(a+b)R_1)\\
=(ab+bc+ca)^3$$
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