Question
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$

Answer

$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&-\text{c}&-\text{b}\\\text{b}&\text{a}+\text{b}+\text{c}&-\text{a}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying C1 → C1 + C2 + C3]
$=\begin{vmatrix}\text{a}+\text{b}&\text{a}+\text{b}&-(\text{a}+\text{b})\\\text{b}+\text{c}&\text{b}+\text{c}&\text{b}+\text{c}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying R1 → R1 + R2 and R2 → R2 + R3]
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}1&1&-1\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
[Taking out common factor from R1 and R2]
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}0&0&-2\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying R1 → R1 - R2]
$=(\text{a}+\text{b})(\text{b}+\text{c})\{(-2)(-\text{a}-\text{c})\}$ [Expanding along R1]
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
$=\text{R.H.S}$
Hence proved.

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