The value of $\begin{bmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{bmatrix},$ is

$(a-b)(b-c)(c-a)$

The correct answer is option (1) : $(a-b)(b-c)(c-a)$

We have,

$Δ=\begin{bmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{bmatrix}$

$⇒Δ=\begin{bmatrix}1 & a & a^2\\0 & b-a & b^2-a^2\\0 & c -a & c^2-a^2\end{bmatrix}$             $\begin{matrix}Applying \, R_2→R_2-R_1\\and\, R_3→ R_3- R_1\end{matrix}$

$⇒Δ = (b-a)(c-a)\begin{bmatrix}1 & a & a^2\\0 & 1 & b+a\\0 & 1 & c+a\end{bmatrix}$             $\begin{matrix}\text{Taking (b-a) and (c-a)}\\\text{common from } R_2 \, and \, R_3\\ \text{respectively}\end{matrix}$

$⇒Δ= -(b-a) (c-a) (c-b) = (a-b) (b-c) (c-a)$