\[\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}=(a-b)(b-c)(c-a)\]
proof
\[\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}\]
applying column transformation \[ c_2-c_1\] and \[ c_3-c_2\]
we get \[ \begin{vmatrix}1&0&0\\a&b-a&c-b\\a^2&b^2-a^2&c^2-b^2\end{vmatrix}\]
=\[ \begin{vmatrix}1&0&0\\a&b-a&c-b\\a^2&(b-a)(b+a)&(c-b)(c+b)\end{vmatrix}\]
Taking \[ (b-a)\] common from \[c_2\] and \[c-b\] from \[c_3\] determinant reduces to
\[(b-a)(c-b)\begin{vmatrix}1&0&0\\a&1&1\\a^2&b+a&b+c\end{vmatrix}\]
Expanding the determinant through row 1 gives \[ (a-b)(b-c)(c-a)\]
