Question

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

Answer 1

$\Delta =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|R_2=R_2-R_1\Rightarrow \Delta =\left|\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 1& c& c^2\end{array}\right|R_3=R_3-R_1\Rightarrow \Delta =\left|\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\end{array}\right|$

Expanding along $C_1$

$\Rightarrow \Delta =1\left\{\right(b-a\left)\right(c^2-a^2)-(c-a\left)\right(b^2-a^2\left)\right\}+0+0\Rightarrow \Delta =(b-a)(c-a)(c+a)-(c-a)(b-a)(b+a)\Rightarrow \Delta =(b-a)(c-a)(c+a-b-a)\Rightarrow \Delta =(b-a)(c-a)(c-b)\Rightarrow \Delta =(b-c)(c-a)(a-b)$