Question

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

• A. $a+b+c$
• B. $(a+b)(b+c)(c+a)$
• C. $a^2+b^2+c^2$
• D. $(a-b)(b-c)(c-a)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $a^2+b^2+c^2$
C $(a-b)(b-c)(c-a)$
D $a+b+c$

$\Delta =\left|\begin{array}{ccc}{a}^2& a^2-(b-c{)}^2& bc\\ b^2& b^2-(c-a{)}^2& ca\\ c^2& c^2-(a-b{)}^2& ab\end{array}\right|=\left|\begin{array}{ccc}{a}^2& a^2& bc\\ b^2& b^2& ca\\ c^2& c^2& ab\end{array}\right|+\left|\begin{array}{ccc}{a}^2& (b-c{)}^2& bc\\ b^2& (c-a{)}^2& ca\\ c^2& (a-b{)}^2& ab\end{array}\right|\Delta =\left|\begin{array}{ccc}{a}^2& (b-c{)}^2& bc\\ b^2& (c-a{)}^2& ca\\ c^2& (a-b{)}^2& ab\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Delta =\left|\begin{array}{ccc}{a}^2& (b-c{)}^2& bc\\ b^2-a^2& (c-a{)}^2-(b-c{)}^2& ca-bc\\ c^2-a^2& (a-b{)}^2-(b-c{)}^2& ab-bc\end{array}\right|=\left|\begin{array}{ccc}{a}^2& (b-c{)}^2& bc\\ b^2-a^2& a^2-b^2-2c(a-b)& c(a-b)\\ c^2-a^2& a^2-c^2-2b(a-c)& b(a-c)\end{array}\right|=(b-a)(c-a)\left|\begin{array}{ccc}{a}^2& (b-c{)}^2& bc\\ b+a& b+a-2c& c\\ c+a& a+c-2b& b\end{array}\right|=(b-a)(c-a)(a^2(b(b+a-2c)-c(a+c-2b))-(b-c{)}^2(b(b+a)-c(c+a))+bc((b+a)(a+c-2b)-(c+a)(b+a-2c)))=(b-a)(c-a)(b-c)(a+b+c)(a^2+b^2+c^2)$