Question

Without expending the determinant,prove that $\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$

Answer 1

Consider the following question.

$\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$

$\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|=\frac{1}{abc}\left|\begin{array}{ccc}{a}^2& a^3& abc\\ b^2& b^3& bca\\ c^2& c^3& abc\end{array}\right|$

$=\frac{abc}{abc}\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& a^3\\ 1& c^2& a^3\end{array}\right|$

$=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& a^3\\ 1& c^2& a^3\end{array}\right|$

Hence, this is the required answer.