Δ = $\left|\begin{array}{ccc}1& 1& 1\\ \mathit{a}& \mathit{b}& \mathit{c}\\ {\mathit{a}}^3& {\mathit{b}}^3& {\mathit{c}}^3\end{array}\right|$
Applying ${\mathit{C}}_1$ → ${\mathit{C}}_1-{\mathit{C}}_3$ and ${\mathit{C}}_2$ → ${\mathit{C}}_2-{\mathit{C}}_3$, we have:
Δ = $\left|\begin{array}{ccc}1-1& 1-1& 1\\ \mathit{a}-\mathit{c}& \mathit{b}-\mathit{c}& \mathit{c}\\ {\mathit{a}}^3-{\mathit{c}}^3& {\mathit{b}}^3-{\mathit{c}}^3& {\mathit{c}}^3\end{array}\right|$
=
= (c - a) (b - c)
Applying ${\mathit{C}}_1$ → ${\mathit{C}}_1+{\mathit{C}}_2$, we have:
Δ = (c - a) (b - c)
= (b - c) (c - a) (a - b)
= (a - b) (b - c) (c - a) (a + b + c)
Expanding along ${\mathit{C}}_1$, we have:
Δ = (a - b) (b - c) (c - a) (a + b + c) - 1 $\left|\begin{array}{cc}0& 1\\ 1& \mathit{c}\end{array}\right|$
= (a - b) (b - c) (c - a) (a + b + c)
Hence proved.