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

L.H.S. :$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\{{x^2}}&1&x\\x&{{x^2}}&1\end{array}} \right|$

Applying ${C_1} \to {C_1} + {C_2} + {C_3}$,

we get
$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + x + {x^2}}&x&{{x^2}}\\{1 + x + {x^2}}&1&x\\{1 + x + {x^2}}&{{x^2}}&1\end{array}} \right|$

Taking $(1 + x + {x^2})$ common from ${C_1}$,

we get
$= (1 + x + {x^2})\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\1&1&x\\1&{{x^2}}&1\end{array}} \right|$

Applying ${C_1} \to {C_1} - {C_2}$,

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

Taking $(1 - x)$ common from ${C_1}$,

we get
$= (1 + x + {x^2})(1 - x)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\0&1&x\\{1 + x}&{{x^2}}&1\end{array}} \right|$
Expanding along ${C_1},$

we get
$= (1 + x + {x^2})(1 - x)\left[ {1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x\\{{x^2}}&1\end{array}} \right| + (1 + x)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&{{x^2}}\\1&x\end{array}} \right|} \right]$

$= (1 - {x^3})[(1 - {x^3}) + (1 + x)({x^2} - {x^2})] = {(1 - {x^3})^2} =$R.H.S.

Hence, proved.