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

$\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|$
$R_1⟶R_1-R_2,R_2⟶R_2-R_3$
$=\left|\begin{array}{ccc}1-x^2& x-1& x^2-x\\ x^2-x& 1-x^2& x-1\\ x& x^2& 1\end{array}\right|$
$=\left|\begin{array}{ccc}(1-x)\left(1+x\right)& -(1-x)& -x(1-x)\\ -x(1-x)& (1-x)\left(1+x\right)& -(1-x)\\ x& x^2& 1\end{array}\right|$
$=(1-x)(1-x)\left|\begin{array}{ccc}1+x& -1& -x\\ -x& 1+x& -1\\ x& x^2& 1\end{array}\right|$
$C_1⟶C_1+C_2+C_3$
$={(1-x)}^2\left|\begin{array}{ccc}0& -1& -x\\ 0& 1+x& -1\\ 1+x+x^2& x^2& 1\end{array}\right|$
$={(1-x)}^2(1+x+x^2)[1+x(1+x)]$
$={(1-x)}^2(1+x+x^2)(1+x+x^2)$
$={\left[(1-x)(1+x+x^2)\right]}^2$
$={(1-x^3)}^2$ $[\because (1-x^3)=(1-x)(1+x+x^2)]$
hence proved.