Question

Solve $\left|\begin{array}{ccc}-2a& a+b& a+c\\ a+b& -2b& b+c\\ c+a& c+b& -2c\end{array}\right|=4(a+b)(b+c)(c+a)$.

Answer 1

$\begin{array}{c}\Rightarrow R_2\to 2R_2+R_1-26\\ \Rightarrow R_3\to 2R_3+R_1-2c\\ \Rightarrow \left|\begin{array}{c}-2aa+ba+c\\ 0-5b+a3c+a\\ 03b+a-5c+a\end{array}\right|\end{array}$

exp and along $C^1$

$\begin{array}{c}\Rightarrow -2a\left|\begin{array}{c}\left(-5b+a\right)3c+a\\ a+3ba-5c\end{array}\right|\\ \Rightarrow -2a\left|\left(a-5b\right)\left(a-5c\right)-\left(a+3b\right)\left(a+3c\right)\right|\\ \Rightarrow -2a\left(a^2-5ac-5ab+25bc-a^2-3ab-3ac-9bc\right)\\ \Rightarrow -2a\left(-8ac-8ab+16bc\right)\\ \Rightarrow 2a\left(8ac+8ab-8bc-8bc\right)\\ \Rightarrow 2a\left\{8c\left(-b+a\right)+8b\left(a-c\right)\right\}\\ \Rightarrow 2a\times 8\left(ac+ab-bc\right)\\ 16a\left(ac+ab-bc\right).\end{array}$