Solution:
$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}$ [R1, R2, R3 are multiplies by a, b and c respectively, therefore we divide by abc]
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix}$
[Taking abc common from C2]$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
We know that the value of a determinant remains unchanged if its rows and columns are interchanged. so,
$\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $
$=-\triangle_1$
$\triangle_1+\triangle_2=0$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.