Question

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

Answer 1

$\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|=\left|\begin{array}{ccc}1+x+x^2& x& x^2\\ 1+x+x^2& 1& x\\ 1+x+x^2& x^2& 1\end{array}\right|$ (using $C_1\to C_1+C_2+C_3$)

Take out $(1+x+x^2)$ common from $C_1$, we get
$=(1+x+x^2)\left|\begin{array}{ccc}1& x& x^2\\ 1& 1& x\\ 1& x^2& 1\end{array}\right|=(1+x+x^2)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1-x& x-x^2\\ 0& x^2-x& 1-x^2\end{array}\right|$ (using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$)

$=(1+x+x^2)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1-x& x(1-x)\\ 0& x(x-1)& 1-x^2\end{array}\right|$

Take out (1-x) common from$R_2$ and same from $R_3$, we get
$=(1+x+x^2)(1-x)(1-x)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& x\\ 0& -x& 1+x\end{array}\right|$

Expanding along $C_1$, we get
$=(1+x+x^2)(1-x)(1-x)\left[\right(1\times 1+x)-(-x\left)\right(x\left)\right]=(1+x+x^2)(1-x)(1-x)(1+x+x^2)=\left[\right(1-x^3\left)\right(1-x^3\left)\right]=(1-x^3{)}^2=RHS[\because 1-x^3=(1-x)(1+x+x^2)]$