Question

Use proportion of determinants to solve
$\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|=(a-b)(b-c)(c-a)(ab+bc+ca)$

Answer 1

Consider the following question

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

Applying $R_2\to R_2-R_1$ $R_3\to R_3-R_1$

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

$\left|\begin{array}{ccc}{a}^2& a^3& 1\\ b^2& b^3& 1\\ c^2& c^3& 1\end{array}\right|=\left|\begin{array}{ccc}{a}^2& a^3& 1\\ b^2-a^2& b^3-a^3& 0\\ c^2-a^2& c^3-a^3& 0\end{array}\right|$

$=(b^2-a^2)(c^3-a^3)-(b^3-a^3)(c^2-a^2)$

$=(b-c)(b+c)(c-a)(c^2+a^2+ac)-(b^2+a^2+ab)(a+b)(c-a)(c+a)$

$=(c-a)(b-a)((b-c)(b+c)(c^2+a^2+ac)-(b^2+a^2+ab)(a+b)(c+a))$

$=(c-a)(b-a)(a-c)(ab+bc+ca)$

Hence, this is the required answer.