Question

Prove that :

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

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{LHS}=\mathrm{\Delta }=\left|ab-cc-b \\ a-cbc-a \\ a-bb-ac\right| \\ \mathrm{\Delta }=\left|a0c-b+a \\ a-cb+c-a0 \\ a-bb+c-ac+a-b\right|\left[\mathrm{Applying}{\mathrm{C}}_2\to {\mathrm{C}}_2+{\mathrm{C}}_3\mathrm{and}{\mathrm{C}}_3\to {\mathrm{C}}_1+{\mathrm{C}}_3\right] \\ =\left(b+c-a\right)\left(c+a-b\right)\left|a01 \\ a-c10 \\ a-b11\right|\left[\mathrm{Taking}\mathrm{out}\mathrm{common}\mathrm{factor}\mathrm{from}{\mathrm{C}}_2\mathrm{and}{\mathrm{C}}_3\right] \\ =\left(b+c-a\right)\left(c+a-b\right)\left\{\left(a\times \left|10 \\ 11\right|\right)+\left(1\times \left|a-c1 \\ a-b1\right|\right)\right\}\left[\mathrm{Expanding}\mathrm{along}{R}_1\right] \\ =\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right) \\ =\mathrm{RHS} \end{gathered}$$