# Prove that:

Question:

Prove that:

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

Solution:

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

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

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

$=2(\mathrm{a}+\mathrm{b}+\mathrm{c}) \mid 1 \quad a \quad b$

$1 \quad b+c+2 a \quad b$

$\begin{array}{lll}1 & a & c+a+2 b\end{array}$ $\left[\right.$ Taking out $2(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common from $\left.\mathrm{C}_{1}\right]$

$\Delta=2(\mathrm{a}+\mathrm{b}+\mathrm{c}) \mid 1 \quad a \quad b$

$\begin{array}{ccc}0 & b+c+a & 0 & \\ 0 & -b-c-a & c+a+b \mid & {\left[\text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\right]}\end{array}$

$=2(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}+\mathrm{c}) \mid 1 \quad a \quad b \quad b$

$\begin{array}{ccc}0 & 1 & 0 \\ 0 & -1 & 1 \mid\end{array}$ [Taking out $(a+b+c)$ common from $R_{2}$ and $R_{3}$ ]

$=2(a+b+c)^{3}\{1(1-0)\} \quad\left[\right.$ Expanding along $\left.C_{1}\right]$

$=2(a+b+c)^{3}$

$=$ RHS