Question

Prove 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

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=\left|\begin{array}{ccc}0& a-b& a^2-b^2\\ 0& b-c& b^2-c^2\\ 1& c& c^2\end{array}\right|=\left|\begin{array}{ccc}0& a-b& (a-b)(a+b)\\ 0& b-c& (a-c)(b+c)\\ 1& b& c^2\end{array}\right|Takingoutcommon(a-b)\&(b-c)from{R}_{1}and{R}_{2}respevtively=(a-b)(a-c)\left|\begin{array}{ccc}0& 1& a+b\\ 0& 1& b+c\\ 1& c& c^2\end{array}\right|Expandingalong{c}_1=(a-b)(b-c)[b+c-a-b]=(a-b)(b-c)(c-a)$