Using properties of determinants prove the following: \begin{vmatrix} 1 & x & x^2 \ x^2 & 1 & x \ x & x^2 & 1 \end{vmatrix} = (1-x^3)^2

First take LHS:
C1 → C1+C2+C3 = \begin{vmatrix} 1+x+x^2 & x & x^2 \ 1+x+x^2 & 1 & x \ 1+x+x^2 & x^2 & 1 \end{vmatrix}
Take out (1+x+x^2) from C1 = (1+x+x^2)\begin{vmatrix} 1 & x & x^2 \ 1 & 1 & x \ 1 & x^2 & 1 \end{vmatrix}
New R2 → R2-R1; R3 → R3-R1 = (1+x+x^2)\begin{vmatrix} 1 & x & x^2 \ 0 & 1-x & x-x^2 \ 0 & x^2-x & 1-x^2 \end{vmatrix}
Expand with C1 = (1+x+x^2)\begin{vmatrix} 1-x & x-x^2 \ x^2-x & 1-x^2 \end{vmatrix}
Take out 1 - x from C1 and same from C2 = (1+x+x^2)(1-x)^2\begin{vmatrix} 1 & x \ x & 1+x \end{vmatrix}
= (1+x+x^2)(1-x)^2(1+x-x^2) = (1+x+x^2)(1-x)^2
= (1-x^3)^2 = RHS
Hence proved.