$\Delta=\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$

Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:

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

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

$=\left(1-x^{3}\right)(1-x)\left|\begin{array}{ccc}1 & 0 & 0 \\ x^{2} & 1+x & x \\ x & -x & 1\end{array}\right|$

Expanding along $R_{1},$ we have:

$\Delta=\left.\left(1-x^{3}\right)(1-x)(1)\right|_{-x} ^{1+x} \quad 1 |$

$=\left(1-x^{3}\right)(1-x)\left(1+x+x^{2}\right)$

$=\left(1-x^{3}\right)\left(1-x^{3}\right)$

$=\left(1-x^{3}\right)^{2}$

Hence, the given result is proved.