$\left| {\begin{array}{cccccccccccccccccccc}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right|$

We have $\left| {\begin{array}{cccccccccccccccccccc}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right|$

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

$= (a + b + c)\left| {\begin{array}{cccccccccccccccccccc}1&1&1\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right|$
[taking $(a + b + c)$

common from the first row
$= (a + b + c)\left| {\begin{array}{cccccccccccccccccccc}0&0&1\\0&{ - (a + b + c)}&{2b}\\{(a + b + c)}&{(a + b + c)}&{(c - a - b)}\end{array}} \right|$

and $\left. {{C_2} \to {C_2} - {C_3}} \right]$
Expanding along ${R_1}$,

$= (a + b + c)\left[ {1\{ 0 + (a + b + {c^2}\} ]} \right.$

$= (a + b + c)\left[ {{{(a + b + c)}^2}} \right]$
$= {(a + b + c)^3}$