Question

$\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|={\left(1-x^3\right)}^2$

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right| \\ =1\left|\begin{array}{cc}1& x\\ x^2& 1\end{array}\right|-x\left|\begin{array}{cc}{x}^2& x\\ x& 1\end{array}\right|+x^2\left|\begin{array}{cc}{x}^2& 1\\ x& x^2\end{array}\right|\left[\mathrm{Expanding}\right] \\ =(1-x^3)-x(x^2-x^2)+x^2(x^4-x) \\ =1-x^3+x^6-x^3 \\ =1-2x^3+x^6 \\ =(1-x^3{)}^2 \end{gathered}$$