$\left|\begin{array}{ccc}1& \mathit{x}& {\mathit{x}}^2\\ {\mathit{x}}^2& 1& \mathit{x}\\ \mathit{x}& {\mathit{x}}^2& 1\end{array}\right|$
Applying ${\mathit{R}}_1$ → ${\mathit{R}}_1+{\mathit{R}}_2+{\mathit{R}}_3$, we have:
Δ =
= 1 + x + ${\mathit{x}}^2$ $\left|\begin{array}{ccc}1& 1& 1\\ {\mathit{x}}^2& 1& \mathit{x}\\ \mathit{x}& {\mathit{x}}^2& 1\end{array}\right|$
Applying ${\mathit{C}}_2$ → ${\mathit{C}}_2-{\mathit{C}}_1$ and ${\mathit{C}}_3$ → ${\mathit{C}}_3-{\mathit{C}}_1$, we have:
Δ = 1 + x + ${\mathit{x}}^2$ $\left|\begin{array}{ccc}1& 0& 0\\ {\mathit{x}}^2& 1-{\mathit{x}}^2& \mathit{x}-{\mathit{x}}^2\\ \mathit{x}& {\mathit{x}}^2-\mathit{x}& 1-\mathit{x}\end{array}\right|$
= (1 + x + ${\mathit{x}}^2$) (1 - x) (1 - x) $\left|\begin{array}{ccc}1& 0& 0\\ {\mathit{x}}^2& 1+\mathit{x}& \mathit{x}\\ \mathit{x}& -\mathit{x}& 1\end{array}\right|$
= (1 - ${\mathit{x}}^3$) (1 - x) $\left|\begin{array}{ccc}1& 0& 0\\ {\mathit{x}}^2& 1+\mathit{x}& \mathit{x}\\ \mathit{x}& -\mathit{x}& 1\end{array}\right|$
Expanding along ${\mathit{R}}_1$, we have:
Δ = (1 - ${\mathit{x}}^3$) (1 - x) (1) $\left|\begin{array}{cc}1+\mathit{x}& \mathit{x}\\ -\mathit{x}& 1\end{array}\right|$
= (1 - ${\mathit{x}}^3$) (1 - x) (1 + x + ${\mathit{x}}^2$)
= (1 - ${\mathit{x}}^3$) (1 - ${\mathit{x}}^2$)
= ${\left(1-{\mathit{x}}^3\right)}^2$
Hence proved.