Question

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

• A. $a^2+b^2+c^2$
• B. (a+b) (b+c) (c+a)
• C. (a-b) (b-c) (c-a)
• D. None of these

Answer 1

The correct option is C (a-b) (b-c) (c-a)

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=\left|\begin{array}{ccc}0& a-b& a^2-b^2\\ 0& b-c& b^2-c^2\\ 1& c& c^2\end{array}\right|,by\begin{array}{c}{R}_1\to R_1-R_2\\ R_2\to R_2-R_3\end{array}$
$=(a-b)(b-c)\left|\begin{array}{ccc}0& 1& a+b\\ 0& 1& b+c\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)\left|\begin{array}{ccc}0& 0& a-c\\ 0& 1& b+c\\ 1& c& c^2\end{array}\right|,by{R}_1\to R_1-R_2=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& b+c\\ 1& c& c^2\end{array}\right|$
= (a-b) (b-c) (a-c). (-1) = (a-b) (b-c) (c-a).