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Question

# Prove that : $\left|\begin{array}{ccc}a& b-c& c-b\\ a-c& b& c-a\\ a-b& b-a& c\end{array}\right|=\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)$

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Solution

## $\mathrm{Let}\mathrm{LHS}=\mathrm{\Delta }=\left|ab-cc-b\phantom{\rule{0ex}{0ex}}a-cbc-a\phantom{\rule{0ex}{0ex}}a-bb-ac\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{\Delta }=\left|a0c-b+a\phantom{\rule{0ex}{0ex}}a-cb+c-a0\phantom{\rule{0ex}{0ex}}a-bb+c-ac+a-b\right|\left[\mathrm{Applying}{\mathrm{C}}_{2}\to {\mathrm{C}}_{2}+{\mathrm{C}}_{3}\mathrm{and}{\mathrm{C}}_{3}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(b+c-a\right)\left(c+a-b\right)\left|a01\phantom{\rule{0ex}{0ex}}a-c10\phantom{\rule{0ex}{0ex}}a-b11\right|\left[\mathrm{Taking}\mathrm{out}\mathrm{common}\mathrm{factor}\mathrm{from}{\mathrm{C}}_{2}\mathrm{and}{\mathrm{C}}_{3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(b+c-a\right)\left(c+a-b\right)\left\{\left(a×\left|10\phantom{\rule{0ex}{0ex}}11\right|\right)+\left(1×\left|a-c1\phantom{\rule{0ex}{0ex}}a-b1\right|\right)\right\}\left[\mathrm{Expanding}\mathrm{along}{R}_{1}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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