Question

Using properties of determinants, prove the following :

$\left[\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right]$ $=(a-b)(b-c)(c-a)(bc+ca+ab)$

Answer 1

$L.H.S=\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|$

applying $R_1\to R_1-R_3$ and $R_2\to R_2-R_3$ we obtain

$\Rightarrow (b-c)(a-c)\left|\begin{array}{ccc}1& a+c& -b\\ 1& b+c& -a\\ c& c^2& ab\end{array}\right|$

applying $R_1\to R_1-R_2$ and expanding the determinanat gives

$\Rightarrow (a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 1& 1\\ 1& b+c& -a\\ c& c^2& ab\end{array}\right|$

$\Rightarrow (a-b)(b-c)(c-a)(bc+ca+ab)=R.H.S$