By using properties of determinants,show that:
$\left|\begin{array}{ccc}-a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2}\end{array}\right|=4a^{2}b^{2}c^{2}$

(N/A) Let $\Delta = \left|\begin{array}{ccc}-a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2}\end{array}\right|$.
Taking out common factors $a, b, c$ from $R_{1}, R_{2}$ and $R_{3}$ respectively:
$\Delta = abc \left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|$.
Taking out common factors $a, b, c$ from $C_{1}, C_{2}$ and $C_{3}$ respectively:
$\Delta = a^{2}b^{2}c^{2} \left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$.
Applying $R_{2} \rightarrow R_{2} + R_{1}$ and $R_{3} \rightarrow R_{3} + R_{1}$:
$\Delta = a^{2}b^{2}c^{2} \left|\begin{array}{ccc}-1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0\end{array}\right|$.
Expanding along $C_{1}$:
$\Delta = a^{2}b^{2}c^{2} [(-1)(0 - 4) - 0 + 0] = a^{2}b^{2}c^{2}(4) = 4a^{2}b^{2}c^{2}$.
Hence,the result is proved.