Question

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

Answer 1

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

$=\left|\begin{array}{ccc}{a}^2& a^2-(b+c{)}^2& bc\\ b^2& b^2-(c+a{)}^2& ca\\ c^2& c^2-(a+b{)}^2& ab\end{array}\right|[C_2\to C_2-4C_3]$

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

$=(a+b+c)\left|\begin{array}{ccc}{a}^2-c^2& 2(a-c)& b(c-a)\\ b^2-c^2& 2(b–c)& a(c-b)\\ c^2& c-a-b& ab\end{array}\right|[R_1\to R_1-R_3;R_2\to R_2-R_3]$

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

$=(a+b+c)(a-c)(b-c)\left|\begin{array}{ccc}a-b& 0& a-b\\ b+c& 2& -a\\ c^2& c-a-b& ab\end{array}\right|[R_1\to R_1–R_2]$

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

$=(a+b+c)(a-c)(b-c)(a-b)\left|\begin{array}{ccc}1& 0& 0\\ b+c& 2& -(a+b+c)\\ c^2& c-a-b& ab-c^2\end{array}\right|[C_3\to C_3–C_1]$

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

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