Question

Using properties of determinants prove the following:
$\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|=(1-x^3{)}^2$.

Answer 1

First take LHS:
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}1+x+x^2& x& x^2\\ 1+x+x^2& 1& x\\ 1+x+x^2& x^2& 1\end{array}\right|$
Take out $(1+x+x^2)$ from $C_1$
$=(1+x+x^2)\left|\begin{array}{ccc}1& x& x^2\\ 1& 1& x\\ 1& x^2& 1\end{array}\right|$
New $R_2\to R_2-R_1:R_3\to R_3-R_1$
$=(1+x+x^2)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1-x& x(1-x)\\ 0& x(x-1)& 1-x^2\end{array}\right|$
Expand with $C_1$
$=(1+x+x^2)\left|\begin{array}{cc}1-x& x(1-x)\\ -x(1-x)& 1-x^2\end{array}\right|$
Take out 1 - x from $C_1$ and same from $C_2$
$=(1+x+x^2)(1-x{)}^2\left|\begin{array}{cc}1& x\\ -x& 1+x\end{array}\right|$
$=(1+x+x^2)(1-x{)}^2(1+x+x^2)$
$={(1+x+x^2)(1-x)}^2$
$=(1-x^3{)}^2$ = RHS
Hence proved.