By using properties of determinants, show that

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

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

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

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

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:

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

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

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

Expanding along R1, we have:

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

Hence, the given result is proved.