Let us consider the L.H.S of the given equation. Let $\begin{vmatrix}1&\text{a}&\text{a}^2\\\text{a}^2&1&\text{a}\\\text{a}&\text{a}^2&1\end{vmatrix}$ Applying C1 → C1 + C2 + C3, we have, $=\begin{vmatrix}1+\text{a}+\text{a}^2&\text{a}&\text{a}^2\\1+\text{a}+\text{a}^2&1&\text{a}\\1+\text{a}+\text{a}^2&\text{a}^2&1\end{vmatrix}$ Taking term (1 + a + a2) common, we have, $=(1+\text{a}+\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\1&1&\text{a}\\1&\text{a}^2&1\end{vmatrix}$ Applying R2 → R2 - R1 and R3 → R3 - R1, we have, $=(1+\text{a}+\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\0&1-\text{a}&\text{a}(1-\text{a})\\0&-\text{a}(1-\text{a})&(1-\text{a})(1+\text{a})\end{vmatrix}$ Taking the term (1 - a) common from R2 and R3, we have, $=(1+\text{a}+\text{a}^2)(1-\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\0&1&\text{a}\\0&-\text{a}&(1+\text{a})\end{vmatrix}$ $=(1+\text{a}+\text{a}^2)(1-\text{a}^2)(1+\text{a}+\text{a}^2)$ $=(1+\text{a}+\text{a}^2)(1-\text{a}^2)$ $=\big[(1+\text{a}+\text{a}^2)(1-\text{a}^2)\big]^2$ $=\big[(\text{a}^3-1)\big]^2$ $=\text{R.H.S}$
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