# Prove that:

Question:

Prove that:

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

Solution:

Let LHS $=\Delta=\mid a-b-c \quad 2 a \quad 2 a$

$\begin{array}{lll}2 b & b-c-a & 2 b\end{array}$

$\begin{array}{lll}2 c & 2 c & c-a-b \mid\end{array}$

$\Rightarrow \Delta=\mid a+b+c \quad a+b+c \quad a+b+c$

$\begin{array}{ccc}2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \mid\end{array}$

$\left[\right.$ Applying $\left.\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}\right]$

$=(a+b+c) \mid \quad 1 \quad 1 \quad 1$

$\begin{array}{ccc}2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \mid\end{array}$

$\begin{array}{llllllllll}=(a+b+c) & 0 & 1 & 1 & b+c+a & b-c-a & 2 b & 0 & 2 c & c-a-b \mid & {\left[\mathrm{Applying} \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}\right]}\end{array}$

$=(a+b+c)\{(a+b+c) \times \mid 1 \quad 1$

$\begin{array}{ll}2 c & c-a-b \mid\}\end{array}$ [Expanding along $\mathrm{C}_{1}$ ]

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

$=R H S$