$\begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix}$ = abc $\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix}$ [Taking out a, b, and c common from $R_1, R_2$, and $R_3$ respectively] = $a^2b^2c^2$ $\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}$ [Taking out a, b, and c common from $C_1, C_2$, and $C_3$ respectively] = $a^2b^2c^2$ $\begin{vmatrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{vmatrix}$ [Applying $R_2$ → $R_2 + R_1$ and $R_3$ → $R_3 + R_1$] = $a^2b^2c^2$ [(-1) (0 × 0 – 2 × 2)] = $a^2b^2c^2$ [- (0 – 4)] = 4 $a^2b^2c^2$ Hence proved.