Question

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

• A. $(a-b)(b-c)(c-a)$
• B. $(a+b)(c-a)$
• C. $(a+b+c{)}^2$
• D. $2(a+b+c{)}^2$

Answer 1

The correct option is A $(a-b)(b-c)(c-a)$

$\begin{array}{c}Letthegivendeterminantbe\Delta .Then\\ \Delta =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\\ =\left|\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\end{array}\right|\left[applying{R}_2\to \left(R_2-R_1\right)and{R}_3\to \left(R_3-R_1\right)\right]\\ =\left(b-a\right)\left(c-a\right).\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b-a\\ 0& 1& c+a\end{array}\right|\\ \left[taking\left(b-a\right)commonfrom{R}_{2}and\left(c-a\right)commonfrom{R}_3\right]\\ =\left(b-a\right)\left(c-a\right)\times 1.\left|\begin{array}{cc}1& b+a\\ 1& c+a\end{array}\right|\left[andedby{C}_1\right]\\ =\left(b-a\right)\left(c-a\right)\left\{\left(c+a\right)-\left(b+a\right)\right\}\\ =\left(b-a\right)\left(c-a\right)\left(c-b\right)=\left(a-b\right)\left(b-c\right)\left(c-a\right).\\ Hence,\\ \Delta =\left(a-b\right)\left(b-c\right)\left(c-a\right).\end{array}$

So, option $A$ is correct answer.