Question
$\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}=4\text{abc}$

Answer

$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}$ [Applying R1 → R1 - (R2 + R3)]
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}-\text{b}&0\\\text{c}&0&\text{a}+\text{b}-\text{c}\end{vmatrix}$ [Applying C2 → C2 - C1 and C3 → C3 - C1]
$=0\begin{vmatrix}\text{c}+\text{a}-\text{b}&0\\0&\text{a}+\text{b}-\text{c}\end{vmatrix}-(-2\text{c})\begin{vmatrix}\text{b}&0\\\text{c}&\text{a}+\text{b}-\text{c}\end{vmatrix}-2\text{b}\begin{vmatrix}\text{b}&\text{c}+\text{a}-\text{b}\\\text{c}&0\end{vmatrix}$
$=2\text{c}[\text{b}(\text{a}+\text{b}-\text{c})-0]-2\text{b}[0-\text{c}(\text{c}+\text{a}-\text{b})]$
$=2\text{bc}[\text{a}+\text{b}-\text{c}]-2\text{bc}[\text{b}-\text{c}-\text{a}]$
$=2\text{bc}[(\text{a}+\text{b}-\text{c})-(\text{b}-\text{c}-\text{a})]$
$=4\text{abc}$
$=\text{R.H.S}$

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