$\left|\begin{array}{ccc}-{\mathit{a}}^2& \mathit{a}\mathit{b}& \mathit{a}\mathit{c}\\ \mathit{b}\mathit{a}& -{\mathit{b}}^2& \mathit{b}\mathit{c}\\ \mathit{c}\mathit{a}& \mathit{c}\mathit{b}& -{\mathit{c}}^2\end{array}\right|$
= abc $\left|\begin{array}{ccc}-\mathit{a}& \mathit{b}& \mathit{c}\\ \mathit{a}& -\mathit{b}& \mathit{c}\\ \mathit{a}& \mathit{b}& -\mathit{c}\end{array}\right|$
[Taking out a, b, and c common from ${\mathit{R}}_1,{\mathit{R}}_2$, and ${\mathit{R}}_3$ respectively]
= ${\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2$ $\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$
[Taking out a, b, and c common from ${\mathit{C}}_1,{\mathit{C}}_2$, and ${\mathit{C}}_3$ respectively]
= ${\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2$ $\left|\begin{array}{ccc}-1& 1& 1\\ 0& 0& 2\\ 0& 2& 0\end{array}\right|$ [Applying ${\mathit{R}}_2$ → ${\mathit{R}}_2+{\mathit{R}}_1$ and ${\mathit{R}}_3$ → ${\mathit{R}}_3+{\mathit{R}}_1$]
= ${\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2$ [(-1) (0 × 0 – 2 × 2)]
= ${\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2$ [- (0 – 4)] = 4 ${\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2$
Hence proved.