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Using the property of determinants and without expanding, prove that
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
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$\begin{aligned} \text { L.H.S. } &=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|: \\=& {\left[\begin{array}{ccc}a-b+b-c+c-a & b-c & c-a \\ b-c+c-a+a-b & c-a & a-b \\ c-a+(a-b)+(b-c) & a-b & b-c\end{array}\right] }\end{aligned}$
$\left(C_1+C_2+C_3 \Rightarrow C_1\right)$
$=\left|\begin{array}{lll}0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c\end{array}\right|=0 \quad\left[\right.$ as $\left._1=0\right]$
$\left(C_1+C_2+C_3 \Rightarrow C_1\right)$
$=\left|\begin{array}{lll}0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c\end{array}\right|=0 \quad\left[\right.$ as $\left._1=0\right]$
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