Question

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

• A. $0$
• B. $a^3+b^3+c^3-3abc$
• C. $3abc$
• D. $(a+b+c{)}^3$

Answer 1

The correct option is A $0$

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