Question
Using properties of determinants, prove the following:

$ \begin{vmatrix} a - b -c & 2a & 2a \\ 2b & b- c - a & 2b \\ 2c & 2c & c- a -b \end{vmatrix} = (a + b + c)^{3}$

Answer

$R_{1} \rightarrow R_{1} + R_{2} + R_{3} \Rightarrow \triangle= (a+ b+ c) \begin{vmatrix} 1 & 1 & 1\\ 2b & b- c - a & 2b \\ 2c & 2c & c- a -b \end{vmatrix} $

$C_{2} \rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1} \Rightarrow $

$\triangle = (a + b + c) \begin{vmatrix} 1& 0 & 0 \\ 2b & -(b+ c + a) & 0 \\ 2c & 0 & -(c+ a +b) \end{vmatrix} $

$= (a + b + c)^{3}$

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