# Solve this

Question:

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

Solution:

Let LHS $=\Delta=\mid a+b+c \quad-c \quad-b$

$\begin{array}{lll}-c & a+b+c & -a \\ -b & -a & a+b+c \mid\end{array}$

$=\mid \mathrm{a} \quad-\mathrm{c} \quad-\mathrm{b}$

$\begin{array}{ccc}b & a+b+c & -a \\ c & -a & a+b+c \mid\end{array}$

$=\mid \mathrm{a} \quad-\mathrm{c} \quad-\mathrm{b}$

$\begin{array}{ccc}b & a+b+c & -a \\ c & -a & a+b+c \mid\end{array}$  [Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$ ]

$=\mid a+b \quad a+b \quad-(a+b)$

$\begin{array}{ccc}\mathrm{b}+\mathrm{c} & \mathrm{b}+\mathrm{c} & \mathrm{b}+\mathrm{c} \\ \mathrm{c} & -\mathrm{a} & \mathrm{a}+\mathrm{b}+\mathrm{c} \mid\end{array}$

[Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3}$ ]

$=(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c}) \mid 1 \quad 1 \quad-1$

$\begin{array}{lll}1 & 1 & 1\end{array}$

$c \quad-a \quad a+b+c \mid \quad$ [Taking out common factor from $R_{1}$ and $R_{2}$ ]

$=(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c}) \mid 0 \quad 0 \quad-2$

$\begin{array}{lll}1 & 1 & 1\end{array}$

$\mathrm{c} \quad-\mathrm{a} \quad \mathrm{a}+\mathrm{b}+\mathrm{c} \mid \quad$ [Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ ]

$=(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})\{(-2)(-\mathrm{a}-\mathrm{c})\} \quad$ [Expanding along $\left.\mathrm{R}_{1}\right]$

$=2(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})$

$=\mathrm{RHS}$

Hence proved.