Question

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

Answer 1

Now,

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

[$R_2^{\prime }=R_2-R_1$ and $R_3^{\prime }=R_3-R_1$]

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

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

[$R_3^{\prime }=R_3-R_2$]

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

$=(b-a)(c-a)(c-b)$

$=(a-b)(b-c)(c-a)$.