Question
Prove that:
$\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}=(\text{a}+\text{b}+\text{c})^3$

Answer

$\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}=(\text{a}+\text{b}+\text{c})^3$
$\text{L.H.S}=\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Apply: $R_1 \rightarrow R_1 + R_2 + R_3$
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Take $(a + b + c)$ common from $R_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Apply: $C_2 \rightarrow C_2- C_1, C_3 \rightarrow C_3 - C_1$​​​​​​​
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\2\text{b}&-\text{b}-\text{c}-\text{a}&0\\2\text{c}&2\text{c}&-\text{c}-\text{a}-\text{b} \end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\2\text{b}&\text{b}+\text{c}+\text{a}&0\\2\text{c}&2\text{c}&\text{c}+\text{a}+\text{b} \end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})^3$
$=\text{R.H.S}$

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