Prove that:
$\left|\begin{array}{lll}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{array}\right|=2\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$
Let $\Delta=\mid \begin{array}{lll}a+b & b+c & c+a\end{array}$
$b+c \quad c+a \quad a+b$
$c+a \quad a+b \quad b+c \mid$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get'
$\Delta=\mid \begin{array}{lll}a & b & c\end{array}$
$\begin{array}{lll}b & c & a\end{array}$
$c \quad a \quad b|+| \begin{array}{lll}b & c & a\end{array}$
$\begin{array}{lll}c & a & b\end{array}$
$a \quad b \quad c$
$=\mid \begin{array}{lll}a & b & c\end{array}$
$\begin{array}{lll}b & c & a\end{array}$
$\begin{array}{lll}c & a & b\end{array}|+(-1)| \begin{array}{lll}a & c & b\end{array}$
$\begin{array}{lll}b & a & c\end{array}$
$\begin{array}{ccc}c & b & a\end{array}$ [Applying $\mathrm{C}_{1} \leftrightarrow \mathrm{C}_{3}$ in second determinant to get negative value of the deteminant]
$=\mid \begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}$
$\begin{array}{lll}b & c & a\end{array}$
$\begin{array}{lll}\mathrm{c} & \mathrm{a} & \mathrm{b}\end{array}|+(-1)(-1)| \begin{array}{lll}a & b & c\end{array}$
$\begin{array}{lll}b & c & a\end{array}$
$\begin{array}{ccc}c & a & b\end{array}$ [Applying $\mathrm{C}_{2} \leftrightarrow \mathrm{C}_{3}$ ]
$=2 \mid a b c$
$\begin{array}{lll}b & c & a\end{array}$
$\begin{array}{lll}c & a & b\end{array}=$ RHS
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