Question

Solve: Value of$D=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$ is

Answer 1

Now,

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

[ $C_2^{\prime }=C_2-C_1$ and $C_3^{\prime }=C_3-C_1$]

or, $D=\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^2& b^2-a^2& c^2-a^2\end{array}\right|$

or, $D=(b-a)(c-a)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2& b+a& c+a\end{array}\right|$

or, $D=(b-a)(c-a)(c-b)$ [ Expanding with respect to first row]

or, $D=(a-b)(b-c)(c-a)$.